Answer:
THE STANDARD HEAT OF SOLUTION OF SODIUM HYDROXIDE IN WATER IS -7.68 KJ PER MOLE.
Explanation:
Variables:
Mass of NaOH = 3.25 g
Mass of water = 50 g
Initial temperature of water = 22°C = 22 + 273 K = 295 K
Final temperature of the reaction mixture = 24.8 °C = 24.8 + 273 K = 297.8 K
Assuming that:
1. specific heat of water = 4.184 J/g °C
2. total mass of the reaction mixture = 50 g + 3.25 g = 53.25 g
3. the rise in temperature = (297.8 K - 295 K ) = 2.8 K
4. Molar mass of sodium hydroxide = ( 23 + 16 + 1) = 40 g/mol
5. number of mole of sodium hydroxide = mass / molar mass
n = 3.25 g / 40 g/mol
n = 0.08125 moles
The rise in temperature for the reaction mixture produces how much of heat:
Heat = mass * specific heat * change in temperature
Heat = 53.25 * 4.184 * 2.8
Heat = 623.8344 J of heat.
Equation of reaction:
NaOH + H2O -------> NaOH + H2O + Heat
This is not a reaction but a dissolution as sodium hydroxide is very soluble in water and this reaction is exothermic where heat is given off.
So since 3.25 g having 0.08125 moles produces 623.8344 J of heat, 1 mole of the sodium hydroxide used will produce:
0.08125 mole of sodium hydroxide = 623.8344 J of heat
1 mole of sodium hydroxide = ( 623.8344 / 0.08125 J of heat
= 7677.96 J of heat per mole of sodium hydroxide.
= 7.68 kJ of heat
So therefore, the standard heat of solution of sodium hydroxide in water is -7.68 kJ of heat since its an exothermic reaction.
Amylose is a "polysaccharide" that plants use to store energy. It is made of repeating subunits of C6H10O5. If a particular amylose molecule has 2537 of these subunits, what is its molecular formula? What is its molar mass? What is the empirical formula of amylose?
Answer:
See exaplanation
Explanation:
For this question, we several sub-questions, lets start with the first one:
What is its molecular formula?
For this question we have to multiply the subunit by the number of subunits in the polysaccharide, so:
[tex](C_6H_1_0O_5)*2537=C_1_5_2_2_2H_2_5_3_7_0O_1_2_6_8_5[/tex]
What is its molar mass?
For this question, we have to find the molar mass of 1 subunit and then multiply by the number of subunits:
Atomic masses: C: 12 g/mol H: 1 g/mol O: 16 g/mol
Now we can multiply, the atomic masses by the number of atoms, so:
[tex](12*6)+(10*1)+(16*5)=~162~g/mol[/tex]
If we take into account the number of subunits:
[tex](162~g/mol)*2537=410994~g/mol[/tex]
What is the empirical formula of amylose?
In this case, we have to remember that the empirical formula is the smallest number of atoms. In other words, we have to simplify the formula. Therefore, the smallest formula is the subunit formula: [tex]C_6H_1_0O_5[/tex].
I hope it helps
You decide to focus on pickling cucumbers. The cucumbers that you have are all approximately 12 cm in length and 3 cm in diameter. If the diffusion coefficient for H+ ions in cucumbers is the same as in water, 7 x 10-5 cm2/sec, how long will it take for the protons to reach the center of the cucumbers? Enter your answer in minutes without including the units.
Answer:
0.0036
Explanation:
SEE THE ATTACHMENT BELOW FOR DETAILED EXPLANATION
Classify each amino acid according to whether its side chain is predominantly protonated or deprotonated at a pH of 7.40. The pKa values of the Asp, His, and Lys side chains are 3.65, 6.00, and 10.53, respectively.
Answer and Explanation:
7.40 becomes lower or shorter than the pKa of functional groups or chains of all the 3 amino acids respectively Lys, His, as well as Asp, although since pH provided throughout the statement. However, all of the covalent chains or bonds of the three compounds or acids (amino) will serve a purpose in the pro-toned pattern.
The pH during which everyone's binding sites will mostly keep track of customer for someone's comparison is:
Lys:
[tex]pH > 10.53[/tex]
His:
[tex]pH > 6.00[/tex]
Asp:
[tex]pH > 3.65[/tex]
Suppose the reaction between nitric oxide and bromine proceeds by the following mechanism: step elementary reaction rate constant (g) (g) (g) (g) (g) (g) Suppose also ≫. That is, the first step is much faster than the second.
Write the balanced chemical equation for the overall chemical reaction:
Write the experimentally-observable rate law for the overall chemical reaction.
Note: your answer should not contain the concentrations of any intermediates.
Answer:
Koverall [NO]^2 [Br2]
Balanced chemical reaction equation;
2NO + Br2 ⇄2NOBr
Explanation:
Consider the first step in the reaction;
NO(g) + Br2(g) ⇄ NOBr2(g) fast
The second step is the slower rate determining step
NOBr2(g) + NO(g) ⇄ 2NOBr(g)
Given that k1= [NOBr2]/[NO] [Br2]
k2= [NOBr2] [NO]
The concentration of the intermediate is now;
[NOBr2]= k1[NO][Br2]
It then follows that overall rate of reaction is
Rate= k1k2[NO]^2 [Br2]
Since k1k2=Koverall
Rate= Koverall [NO]^2 [Br2]
Answer:
NOBr_2(g)+NO(g)----> 2NOBr(g)
Explanation:
Answer retrieved from ALEKS
Draw the reaction mechanism the reaction between 2-methylpent-2-ene and chlorine. State the type of reaction. Label the electrophile, intermediate and final product.
Answer:
Explanation:
CH₃C(CH₃)=CH-CH₂-CH₃ ( 2 - methylpent-2-ene)
Cl₂ ⇒ Cl⁺ + Cl⁻
Cl⁺( electrophile)
CH₃C(CH₃)=CH-CH₂-CH₃ + Cl⁺ ( electrophile) ⇒ CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ ( intermediate )
CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ + Cl⁻ ⇒ CH₃CCl(CH₃)-C(Cl)H-CH₂-CH₃ ( final product )
It is electrophilic addition reaction.
How many molecules are in 0.3672 moles of LiBr?
Answer:
2.21×10²³molecules
Explanation:
Hello,
The question requires us to calculate the number of molecules present in a given number of mole.
Numbe of mole = 0.3672 moles
According to mole ratio concept, one mole of any substance is equal to Avogadro's number of particles, atoms or molecules. I.e,
1 mole = Avogadro's number
Avogadro's number = 6.022×10²³ atoms/molecules/particles or ions
1 mole = 6.022×10²³molecules
0.3672 mole = x molecules
x = 0.3672 × 6.022×10²³
X = 2.21×10²³molecules
Therefore, 0.3672moles of LiBr contains 2.21×10²³molecules
Consider four beakers labeled A, B, C, and D, each containing an aqueous solution and a solid piece of metal. Identity the beakers in which a chemical reaction will occur and those in which no reaction will occur. Drag each item to the correct bin.
1. Mn(s) Ca(NO3)2(aq)
2. KOH(aq) Fe(s)
3. Pt(NO3)2(aq) Cu(s)
4. Cr(s) H2SO4(aq)
A. Reaction
B. Non-reaction
Consider the following data for five hypothetical element: Q, W, X, Y, and Z. Rank the elements from most reactive to least reactive.
Combination Observation of reaction
Q + W+ Reaction occurs
X + Z+ No reaction
W + Z+ Reaction occurs
Q+ + Y Reaction occurs
Place the element symbol from most to least reactive.
1. Q
2. W
3. X
4. Y
5. Z
A. Most reactive
B. Least reactive
Use the following reactions to arrange the elements A, B, C, and D in order of their rodox reactivity from most reactive to least reactive. Rank from most reactive to least reactive.
1. A + D+ righ tarrow A+ + D
2. C+ + D righ tarrow C + D+
3. B+ + D righ tarrow B + D+
4. B + C+ righ tarrow B+ + C
A. Most reactive
B. Least reactive
Answer:
A Reaction
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
B Non Reaction
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
Y > Q > W > Z > X
Explanation:
The first question is whether a reaction will occur base on the chemical equation below.
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
Firstly, some element are more reactive than others , base on this criteria element can be arranged base on it reactivity .
1. Mn(s) + Ca(NO₃)₂(aq)
This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.
2. KOH(aq) + Fe(s)
The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.
3. Pt(NO₃)₂(aq) + Cu(s)
Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.
4. Cr(s) + H₂SO₄(aq)
Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.
Base on the reaction
Q + W+ Reaction occurs
Since the reaction occurred element Q is more reactive as it displace element w from it compound.
X + Z+ No reaction
No reaction occurred because element x is less reactive than z therefore, it cannot displace z from it compound.
W + Z+ Reaction occurs
Element w is more reactive than z as it displaces z form it compound.
Q+ + Y Reaction occurs
Element Y is more reactive than element Q as it displaces Q from it compound.
Therefore, the order of reactivity from the most reactive to the least reactive will be Y > Q > W > Z > X
A. The beakers in which a chemical reaction will occur:
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
B. The beakers in which there is no reaction:
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
C. The elements from most reactive to least reactive is:
Y > Q > W > Z > X
Solving for each part:A.
1. Mn(s) + Ca(NO₃)₂(aq)
2. KOH(aq) + Fe(s)
3. Pt(NO₃)₂(aq) + Cu(s)
4. Cr(s) + H₂SO₄(aq)
Elements can be arranged on the basis of reactivity:
1. Mn(s) + Ca(NO₃)₂(aq)
This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.
2. KOH(aq) + Fe(s)
The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.
3. Pt(NO₃)₂(aq) + Cu(s)
Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.
4. Cr(s) + H₂SO₄(aq)
Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.
According to reactions:
Q + W→ Reaction occurs
Since the reaction occurred element Q is more reactive as it displace element w from it compound.
X + Z →No reaction
No reaction occurred because element X is less reactive than Z therefore, it cannot displace z from it compound.
W + Z→ Reaction occurs
Element W is more reactive than Z as it displaces Z form it compound.
Q + Y →Reaction occurs
Element Y is more reactive than element Q as it displaces Q from it compound.
Thus, the order of reactivity from the most reactive to the least reactive will be: Y > Q > W > Z > X
Find more information about Reactivity series here:
brainly.com/question/17469010
If 200 mL of 3M CaCO3 is diluted to 250 mL, what is the new molarity?
Answer: The new molarity is 2.4 M
Explanation:
According to the dilution law,
[tex]C_1V_1=C_2V_2[/tex]
where,
[tex]C_1[/tex] = concentration of pure solution = 3 M
[tex]V_1[/tex] = volume of pure solution = 200 ml
[tex]C_2[/tex] = concentration of diluted solution= ?
[tex]V_2[/tex] = volume of diluted solution= 250 ml
Putting in the values:
[tex]3M\times 200ml=M_2\times 250ml[/tex]
[tex]M_2=2.4M[/tex]
Thus the new molarity is 2.4 M
Classify each titration curve as representing a strong acid titrated with a strong base, a strong base titrated with a strong acid, a weak acid titrated with a strong base, a weak base titrated with a strong acid, or a polyprotic acid titrated with a strong base. A. Strong acid/ strong base.B. Strong base/ strong acid.C. Weak acid/ strong base.D. Weak base/ strong acid.E. Polyprotic acid/ strong base.
Answer:
Acid base titration curves shows the pH at equivalence point
Explanation:
Since the images were not shown, I will proceed to give a general description of the following acid-base titration curves:
In a strong acid-strong base titration, the acid and base will react to form a neutral solution. At the equivalence point of the reaction, hydronium (H+) and hydroxide (OH-) ions will react to form water, leading to a pH of 7.
The titration curve reflects the strengths of the corresponding acid and base. If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular, and the pH shifts less with small additions of titrant near the equivalence point.
Polyprotic acids are able to donate more than one proton per acid molecule, in contrast to monoprotic acids that only donate one proton per molecule. In the titration curve of a polyptotic acid and a strong base, The curve starts at a higher pH than a titration curve of a strong base. There is always a steep climb in pH before the first midpoint. Gradually, the pH increases until it passes the midpoint; Right before the equivalence point there is a very sharp increase in pH.
Answer:
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Explanation:
Fe3O4(s) + CO(g) → FeO(s) + CO2(g) What coefficients will balance the equation? 3, 1, 1, 1 1, 1, 3, 1 3, 3, 1, 1 2, 2, 6, 4
Answer:
1, 1, 3, 1.
Explanation:
When balancing chemical equations, the amount of moles of each element on both sides of the equation should be equal.
Looking at the original equation, we can see that there are 3 mols of iron on the reactants side and 1 on the products. We can simply add a coefficient of '3' to 'FeO' to balance the iron.
For Oxygen, we can see 5 on the reactants side. However, since we already added a coefficient of '3' to 'FeO', this already balanced out the oxygen for us. We have 5 mols in the reactants, and 5 in the products.
Carbon is already balanced on both sides.
Therefore, the final formula is [tex]Fe_{3}O_{4} + CO -> 3FeO + CO_{2}[/tex], with coefficients of 1, 1, 3, and 1.
Answer:
B
Explanation:
Fe3O4(s) + CO(g) →FeO(s) + CO2(g)
Left side
Fe =3
O = 5
C = 1
Right side
Fe =1
O = 3
C = 1
Balance by finding common denominator
We'll make Fe 3 on the left side
Fe3O4(s) + CO(g) →3FeO(s) + CO2(g)
Left side
Fe =3
O = 5
C = 1
Right side
Fe =3
O = 5
C = 1
What is the chemical formula for a compound between Al and F? AlF AlF 3 AlF 2 Al 3F
Answer:
AlF3
Explanation:
Aluminum fluoride (AlF3) is an inorganic compound in which aluminum and fluoride has ionic bond between each other.
Aluminum ( atomic number - 13) has 3 valence electrons and can lose 3 electrons to get a stable configuration while fluorine (atomic number - 9) can gain one electron to make a stable configuration. So in order to attain a stable configuration one aluminum (Al) atom form ionic bond with three fluorine atoms and form AlF3.
Hence, the chemical formula between Al and F is AlF3.
Answer:
Explanation:
mmmmmmmmmmmmmh,mm
Calculate the empirical formula of a compound with the following
percentage composition: 31,8% K; 29,0% C1; 39,2% O
The empirical formula of ethane is CH. Its molar mass is 30 g.mol-
Answer:
The empirical formula for this compound is thus CH2.
Explanation:
The standard free energy change in physiological conditions (G') for the reaction catalyzed by malate dehydrogenase in the citric acid cycle malate + NAD+ oxaloacetate + NADH + H+ is ~+29 kJmol-1 Calculate the actual G' at 37C if Keq' is 1.02 × 10-5
Answer: The actual value of [tex]\Delta G[/tex] is -618 J/mol
Explanation:
Relation of ree energy change and equilibrium constant
[tex]\Delta G=\Delta G^0+2.303\times RT\times \log K_{eq}[/tex]
where,
[tex]\Delta G[/tex] = Free energy change
[tex]\Delta G^o[/tex] = standard free energy change = +29 kJ/mol =
R = universal gas constant
T = temperature = [tex]37^0C=(37+273)K=310K[/tex]
[tex]K_{eq}[/tex] = equilibrium constant = [tex]1.02\times 10^{-5}[/tex]
[tex]\Delta G=+29000J/mol+2.303\times 8.314J/Kmol\times 310K\times \log (1.02\times 10^{-5})[/tex]
[tex]\Delta G=29000J/mol-29619J/mol=-618J/mol[/tex]
The actual value of [tex]\Delta G[/tex] is -618 J/mol
Aluminum metal reacts with sulfuric acid according to the equation:2Al(s) + 3H2SO4(aq) --> Al2(SO4)3(s) + 3H2(g)If 46 g of aluminum reacts with excess sulfuric acid, and 108 g of Al2(SO4)3 are collected, what is the percent yield for the reaction?
Answer:
Percentage yield = 38.37%
Explanation:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃
2 moles of Al contains = (2 × 27) = 54g of Al
1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃
54g of Al produces 342.15g of Al₂(SO₄)₃,
46g of Al will produce x g of Al₂(SO₄)₃
Solve for x
X = (46 × 342.15) / 54
X = 15738.9 / 54
X = 281.46g
Theoretical yield of Al₂(SO₄)₃ = 281.46g
Percentage yield of a substance = (actual yield / theoretical yield) × 100
Actual yield = 108g
Theoretical yield = 281.46
% yield = (108 / 281.46) × 100
% yield = 0.3837 × 100
%yield = 38.37
The percentage yield of Al₂(SO₄)₃ is 38.37
Answer: Percentage yield = 38.37%
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂ from the equation of reaction, 2 moles of Al reacts to produce 1 moles of Al₂(SO₄)₃ .
2 moles of Al contains = (2 × 27) = 54g of Al
1 mole of Al₂(SO₄)₃ contains (1 × 342.15) = 342.15g of Al₂(SO₄)₃
54g of Al produces 342.15g of Al₂(SO₄)₃,
46g of Al will produce x g of Al₂(SO₄)₃
Solve for x
X = (46 × 342.15) / 54
X = 15738.9 / 54
X = 281.46g
Theoretical yield of Al₂(SO₄)₃ = 281.46g
Percentage yield of a substance = (actual yield / theoretical yield) × 100
Actual yield = 108g
Theoretical yield = 281.46
% yield = (108 / 281.46) × 100
% yield = 0.3837 × 100
%yield = 38.37
The percentage yield of Al₂(SO₄)₃ is 38.37
Convert the average temperatures for each collected data point given below from °C to K. Plot the average cell potentials E (y-axis) vs T (x-axis). The plot should be approximately linear. Add a trendline to find the best linear fit and write down the y-intercept and slope (b and m from the linear equation) for the trendline below.
Average Temperature in °C - Average Cell Potential (V)
15 - 0.465
18 - 0.467
21 - 0.468
24 - 4.69
27 - 0.471
30 - 0.472
33 - 0.474
Answer:
Explanation:
The equation of above line , y = 0.0005x+ 0.458
This can be compared with y = mx+c
Hence slope, m = 0.0005 and Y-intercept, c = 0.458
Or it can be plotted manually where straight line has to be drawn touching maximum number of data points. After drawing a straight linear line, we need to take any two points from the straight line and slope is calculated
Slope,
[tex]m = \frac{y_2-y_1}{x_2-x_1}[/tex]
and y -intercept is calculated using extraplotting backwards such that it touches the Y-axis. the point where straight line touches Y-axis is Y-intercept (c).
Plot the average cell potentials E (y-axis) vs T (x-axis). image attached
Atoms of the same element are alike because they must have the same number of
Answer:
They must have the same number of protons.
Explanation:
Protons determine the identity of an element. However, the neutrons can vary, resulting in different masses.
Answer:
Atoms of the same element can have different numbers of neutrons, however. Atoms of the same element (i.e., atoms with the same number of protons) with different numbers of neutrons are called isotopes. Most naturally occurring elements exist as isotopes.
Explanation:
please help me with this!
Answer:
mole fraction of Oxygen = mole of oxygen/ total mole
Explanation:
Can the ph scale be utilized for all acids (Arrhenius, Bronsted-Lowery, and Lewis)? Explain reasoning.
Answer:
No, the pH scale can not be utilized for all types of acids known
Explanation:
The pH scale is a scale that shows the degree of acidity or alkalinity of a substance. The pH scale is graduated from 0-14. The pH scale is mostly applied to Arrhenius acids. Recall that in the Arrhenius definition, an acid produces hydrogen ion in solution while a base produces hydroxide ion in solution. Hence we could easily measure or calculate the pH and pOH of Arrhenius acids. Arrhenius acids apply mostly to acid-base discussion in Aqueous media.
Beyond aqueous media, we can still apply the Brownstead-Lowry and Lewis definitions for acid and base. Some of these substances to which we apply these other definitions of acid and base do not necessarily contain hydrogen or hydroxide ions hence the pH scale can not be utilized in discussing their acid-base properties, hence the answer.
A 38.8 gram piece of metal absorbs 181J as it temperature increases from 25.0 degree celsius to 36.0 degree celsius. What is the specific heat of the metal?
__________________________J/g degree C
Answer: 0.424 J/g°C
Explanation:
For this problem, we would have to manipulate the equaiton for heat, q=mCT. Specific heat is the C in the equation. Since we are looking for specific heat, we manipulate the equation so that it says C=.
[tex]C=\frac{q}{m(deltaT)}[/tex]
*I didn't know how to type in delta so I just wrote the word delta, but pretend you see a Δ.
Now that we have our equation, we can plug in our values and solve.
[tex]C=\frac{181J}{(38.8g)(36-25°C)}[/tex]
*Please ignore the capital A in the equation. It pops up every time I type in the ° sign.
[tex]C=0.424J/g°C[/tex]
A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 93.0 mL. The liquid solvent has a mass of 33.7 g and a density of 0.865 g/mL. Determine the mass of the solid given its density is 3.50 g/mL
Answer:
Mass of solid = 189.141 gram
Explanation:
Given:
Total volume = 93 ml
Mass of liquid = 33.7 gram
Density of liquid = 0.865 g/ml
Density of solid = 3.50 g/ml
Find:
Mass of solid = ?
Computation:
Volume of liquid = Mass of liquid / Density of liquid
Volume of liquid = 33.7 / 0.865
Volume of liquid = 38.9595 ml
Volume of solid = Total volume - Volume of liquid
Volume of solid = 93 - 38.9595
Volume of solid = 54.0405 ml
Mass of solid = Volume of solid × Density of solid
Mass of solid = 54.0405 ml × 3.50 g/ml
Mass of solid = 189.141 gram
2 MnO4−(aq) + 10 Cl−(aq) + 16 H+(aq) → 5 Cl2(g) + 2 Mn2+(aq) + 8 H2O(l) Oxidizing agent: Reducing agent:
Answer:
MnO4 is the oxidizing agent.
Cl is the reducing agent.
Explanation:
To know which is oxidizing agent and which is reducing agent, let us calculate the change in oxidation number of each element in the equation below:
2MnO4−(aq) + 10Cl−(aq) + 16H+(aq) → 5Cl2(g) + 2Mn2+(aq) + 8H2O(l)
Please note:
1. The oxidation state of oxygen is always –2 except in peroxide where it is –1
2. The oxidation state of Hydrogen is always +1 except in hydrides where it is –1
3. The oxidation state of chlorine is always –1
For Mn:
At left hand side
MnO4 = –1
Mn + 4O = –1
O = –2
Mn + (4 x –2) = –1
Mn – 8 = –1
Collect like terms
Mn = –1 + 8
Mn = +7
At the right hand side, Mn is +2.
The oxidation state of Mn changes from +7 to +2.
For Cl:
At the left hand side
Cl = –1
At the right hand side
Cl = 0
The oxidation state of Cl changes from – 1 to 0.
The oxidation state of O and H are unchanged.
Since the oxidation state of Mn changes from +7 to +2 i.e reduced, MnO4 is the oxidizing agent.
Since the oxidation state of Cl changes from – 1 to 0 i.e increase, Cl is the reducing agent.
1) How many kJ are absorbed when 45.2 g of water at 31.3 oC is heated to 76.9 oC? 2) Calculate the total heat released in kcal when 72.1 g water at 25.2 oC is cooled to 0 oC and freezes. 3) How many kilojoules are required to heat 55,500 mg of gold with specific heat = 0.129 J/g oC is heated from 24.6 oC to 123.4 oC? 4) Calculate the heat needed in kcal to change 45.6 g of water at 100 oC to change into steam.
Answer:
1. Q = 8.66 KJ
2. Q = 7.58 Kcal
3. Q = 0.71 KJ
4. Q = 24.31 Kcal
Explanation:
1. The quantity of heat absorbed can be determined by:
Q = mcΔθ
where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of water = 4.2 j/g[tex]^{0}C[/tex] and Δθ is the change in temperature.
= 45.2 × 4.2 × (76.9 - 31.3)
= 8656.704
∴ Q = 8.66 KJ
The quantity of heat absorbed is 8.66 KJ.
2. Q = mcΔθ + mL
Where L is the latent heat of fusion of ice = 334 J.
= m(cΔθ + L)
= 72.1(4.2 × 25.2 + 334)
Q = 31712.464 J
= 7579.466 calories
The total heat released is 7.58 Kcal.
3. Q = mcΔθ
= 55.5 × 0.129 × (123.4 - 24.6)
= 707.3586
The quantity of heat required to increase the temperature of gold is 0.71 KJ.
4. Q = mL
Where: L is the specific latent heat of vaporization = 533 calories.
Q = 45.6 × 533
= 24304.8
The quantity of heat required to change water to steam is 24.31 Kcal.
Kate’s blood volume is 3.8L. Before treatment,if her blood glucose was 170 mg/dL, how many grams of glucose were in her blood?
Answer:
6.460 gm is the correct answer .
Explanation:
Given
Kate’s blood volume=[tex]3.8L[/tex]
blood glucose=[tex]170\ mg/dL[/tex]
As we know that
[tex]1\ dL= 10^{-1}\ litre[/tex]
The [tex]10^{-1}[/tex] includes the blood glucose of the [tex]170\ mg/dL,[/tex]
So [tex]3.8L.[/tex] of the Kate’s blood volume can be determined by
[tex]=\ \frac{170 * 3.8 }{10^{-1} }[/tex]
[tex]=170 * 38 \ mg[/tex]
[tex]=6460\ mg[/tex]
To convert into the gm we have divided by the 1000
[tex]6.460 \ gm[/tex]
Therefore the glucose in her blood=
[tex]6.460 \ gm[/tex]
Answer:
6.460 g
Explanation:
Glucose is a simple carbohydrate molecule which circulates in the blood along with other substances. The amount of glucose is measured to treat the medical conditions of patients which is measured as mg/dL.
1 liter of liquid contains 10 deciliters therefore 1 dL will have 0.1 liter or
1 dL= 10^-1 liter (0.1)
In the given question,
the total volume of blood = 3.8 L
1 dL of blood contains = 170 mg
3.8 L will have = (3.8 x 170)/0.1
=6460 mg
= 6.460 g
Thus, 6.460 g is the correct answer.
PLEAS HELP I HAVE LIMITED TIME!!
Which compound is the limiting reagent?
Select one:
O2
O H20
H2
Cannot be determined.
Answer:
H2 is the limiting reactant.
Explanation:
From the diagram above:
H2 => White ball
O2 => Red ball
Before the reaction
H2 => White ball => 10
O2 => Red ball => 7
After the reaction
H2O => White and red ball => 10
O2 => 2
From the simple illustration above, we can see that all the H2 were used up in the reaction but there are left over of O2.
This simply means that H2 is the limiting reactant as all of it is used up in the reaction while O2 is the excess reactant as there are leftover.
what type of reaction is Cl2(g) + 2Kl(aq) arrow 2KCl(aq) + I2 (s)
Which describes the correct procedure when converting a number from scientific notation to standard notation? If the power of 10 is positive, move the decimal point to the left. If the power of 10 is positive, move the decimal point to the right. If the number being converted is greater than 10, move the decimal point to the left. If the number being converted is greater than 10, move the decimal point to the right.
Answer:
option 2Explanation:
If the power of ten is positive, you move the decimal place to the right when converting a number from scientific to standard notation
For example:
5.6894 * 10^9
10^9 is 1 with 9 zeros
5.6894 * 1000000000
5689400000.
it moved right
If the power of 10 is positive, move the decimal point to the left.
If the power of 10 is positive, move the decimal point to the right.
If the number being converted is greater than 10, move the decimal point to the left.
If the number being converted is greater than 10, move the decimal point to the right.
This is option 2
The correct answer is if the power of 10 is positive, move the decimal point to the right.
What is scientific notation?A frequently-used floating-point system in which integers are expressed as products of a number between 1 and 10 multiplied by a power of 10.
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Tenormin, a member of the group of drugs known as beta-blockers, is used to treat high blood pressure and improve survival after a heart attack. It works by slowing down the heart to reduce its workload. Which atom in Tenormin is the most basic?
Answer:
Oxygen
Explanation:
In the compound tenormin, there are two highly electronegative atoms capable of accepting electrons; oxygen and nitrogen. Oxygen is more electronegative than nitrogen.
However, the oxygen atom in tenormin is bonded to carbon in a carbonyl bond. Recall that the carbonyl bond is polar and the direction of the dipole is towards the oxygen atom. Looking at the structure of tenormin, it is clear that the electron density of the bond tends towards the oxygen atom of the carbonyl group. Electron density is withdrawn from the adjacent nitrogen atom of the amine group via mesomeric and inductive mechanism towards the more electronegative oxygen atom.
On the other side of the structure, there are two oxygen atoms. These oxygen atoms are more electronegative than nitrogen thus they are more basic.
Hence the oxygen atom is the most basic atom in the compound tenormin.
When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-path length cell. For comparison, a 10.0-mL reference sample of 6.74 times 10^-4 M Fe^3+ was treated with HNO_3 and KSCN and diluted to 50.0 mL, The reference was placed in a cell with a 1.00-cm light path. The runoff sample exhibited the same absorbance as the reference when the path length of the runoff cell was 2.41 cm. What was the concentration of iron in Uncle Wilbur's runoff?
Answer:
C = 2.24x10⁻⁴ M
Explanation:
The concentration of iron in Uncle Wilbur's runoff can be calculated using Beer-Lambert law:
[tex] A = \epsilon*C*l [/tex] (1)
Where:
A: is the absorbance of the compound
ε: is the molar absorptivity of the compound
C: is the concentration of the compound
l: is the optical path length
Since the runoff sample exhibited the same absorbance as the reference sample, we can find the concentration using equation (1):
[tex] \epsilon*C_{1}*l_{1} = \epsilon*C_{2}*l_{2} [/tex] (2)
Where:
Subscripst 1 and 2 refer to Uncle Wilbur's runoff and to reference sample, respectively.
l₁ = 2.41 cm
l₂ = 1.00 cm
We can find C₂ as follows:
[tex] C_{2} = \frac{C_{2i}*V_{i}}{V_{f}} [/tex] (3)
Where:
[tex]C_{2i}[/tex]: is the initial concentration of the reference sample = 6.74x10⁻⁴ M
[tex]V_{i}[/tex]: is the initial volume = 10.0 mL
[tex]V_{f}[/tex]: is the final volume = 50.0 mL
[tex] C_{2} = \frac{6.74 \cdot 10^{-4} M*10.0 mL}{50.0 mL} = 1.35 \cdot 10^{-4} M [/tex]
Now, we can find C₁ using equation (2):
[tex] C_{1} = \frac{C_{2}*l_{2}}{l_{1}} = \frac{1.35 \cdot 10^{-4} M*1.00 cm}{2.41 cm} = 5.60 \cdot 10^{-5} M [/tex]
Finally, since the runoff solution was diluted to 100.0 mL, the initial concentration can be calculated using equation (3) for [tex]C_{1i}[/tex]:
[tex]C_{1i} = \frac{C_{1}*V_{f}}{V_{i}} = \frac{5.60 \cdot 10^{-5} M*100.0 mL}{25.0 mL} = 2.24 \cdot 10^{-4} M[/tex]
Therefore, the concentration of iron in Uncle Wilbur's runoff is 2.24x10⁻⁴ M.
I hope it helps you!
A solution consists of 0.27 M MgCl2 and 0.7 M CuCl2. Calculate the concentration of hydroxide ion needed to separate the metal ions. The Ksp of Mg(OH)2 is 6.3x10-10, and the Ksp of Cu(OH)2 is 2.2x10-20. Multiply your answer by 106 and enter that number to 2 decimal places.
Answer:
The concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰
Explanation:
Based on the different Ksp of the hydroxides of the metal ions it is possible to precipitate 1 metal as hydroxide whereas the other still in solution.
Ksp of both hydroxides is:
Mg(OH)₂ ⇄ Mg²⁺ + 2OH⁻
Ksp = [Mg²⁺] [OH⁻]² = 6.3x10⁻¹⁰
Cu(OH)₂ ⇄ Cu²⁺ + 2OH⁻
Ksp = [Cu²⁺] [OH⁻]² = 2.2x10⁻²⁰
Replacing with the concentration of the metals:
[0.27] [OH⁻]² = 6.3x10⁻¹⁰
[OH⁻] = 4.83x10⁻⁵M
That means the precipitation of Mg²⁺ begins when [OH⁻] is 4.83x10⁻⁵M
[0.7] [OH⁻]² = 2.2x10⁻²⁰
[OH⁻] = 1.77x10⁻¹⁰
As the Cu²⁺ needs a low concentration of OH⁻ to begin precipitation,
the concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O What mass of O2 would be required to react completely with 6.87 g of NH3?
Answer:
16.16g of O2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
4NH3 + 5O2 → 4NO + 6H2O
Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 from the balanced equation = 4 x 17 = 68g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 5 x 32 = 160g.
From the balanced equation above,
68g of NH3 reacted with 160g of O2.
Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:
From the balanced equation above,
68g of NH3 reacted with 160g of O2
Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.
Therefore, 16.16g of O2 is needed for the reaction.
Mass of oxygen required for the reaction is 16.16 g.
The equation of the reaction is;
4NH3 + 5O2 → 4NO + 6H2O
Since Number of moles = Mass/molar mass
Molar mass of ammonia = 17 g mol
Number of moles of NH3 reacted = 6.87 g/17 g mol = 0.404 moles
From the reaction equation;
4 moles of ammonia reacts with 5 moles of oxygen
0.404 moles reacts with 0.404 moles × 5 moles/4 moles = 0.505 moles
Mass of oxygen = 0.505 moles × 32 g/mol = 16.16 g
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