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
QUESTION 5
The attraction between the nucleus of the atom and the valence electrons for any given atom is called what?
Answer:
Electrostatic Force of Attraction
Hope this helps ;)
Explanation:
which two compounds are structural isomers of each other
Answer:
Explanation:
Structural isomer is a type of isomer in which molecules with the same molecular formula have different bonding patterns and atomic organization.
For example Pentan-1-ol, pentan-2-ol, and pentan-3-ol are structural isomers that exhibit position isomerism.
Cyclohexane and hex-1-ene are examples of functional group structural isomers.
Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 25 grams of C is formed in 8 minutes. How much (in grams) is formed in 16 minutes
Answer: 45. 78 g is formed in 16 minutes.
Explanation:
Let [tex]A_{o} =40 g[/tex] and [tex]B_{o} =50 g[/tex]
we know that [tex]\alpha =A_{o}\frac{M+N}{M}\, and \, \beta =B_{o}\frac{M+N}{N}[/tex]
According to the question to create [tex]x[/tex] part of the chemical C we will need 2 part of A and one part of B.
Therefore, M = 2 and N = 1.
Now, we can easily solve for the value of α and β.
[tex]\alpha =40\frac{2+1}{2}=60\\\\\, and \, \beta =50\frac{2+1}{1}=150[/tex]
Now, the differential equation must be: [tex]\frac{dX}{dt}=k\left ( \alpha -X \right )\left ( \beta -X \right )[/tex]
I separate the variable and solve the equation
[tex]\int \frac{dx}{\left ( 60-x \right )\left ( 150-x \right )}=\int k\, dt[/tex]
[tex]ln\frac{150-x}{60-x}=90kt+C_{1}[/tex]
By using [tex]X(0) =0[/tex],
[tex]\frac{150-x}{60-x}=Ce^{90k_{o}},C=\frac{5}{2}[/tex]
and using [tex]X (8)=25[/tex] and solving for the value of 'k'
[tex]\Rightarrow \frac{150-25}{60-25}=\frac{5}{2}e^{450k}\\\\\Rightarrow 3.6\times \frac{2}{5}=e^{450k}\\\\\Rightarrow 1.4=e^{450k}[/tex]
Taking 'ln' both side, we get
[tex]\Rightarrow k=7.4716\times 10^{-4}[/tex]
We obtain:[tex]X(t)=\frac{60Ce^{90kt}-150}{Ce^{90kt}-1}[/tex]
Now, for the 16 min
[tex]\Rightarrow X(16)=\frac{150e^{1.07559}-150}{2.5e^{1.0759}-1}[/tex]
[tex]\Rightarrow X(16)=\frac{439.7583-150}{7.3293-1}\\\\\Rightarrow X(16)=\frac{289.7583}{6.3293}[/tex]
[tex]\Rightarrow X(16)=45.78 g[/tex]
Why do heart diseas patient's eat oil instead of fat?
Answer:
lube
Explanation:
Please show work and explain:
The reaction A + B → C is studied similarly to our study of phenolphthalein fading kinetics.
The corresponding rate law is Rate = k[A]m[B]m
In this particular experiment, the concentration of A is sufficiently high that the pseudo-order rate law: Rate = k1[B]n can be written, where k1 = k[A]m.
If the study establishes that the reaction is first order in B (n = 1) and that the pseudo rate constant k1 has the following values at varying concentrations of B, what is the order of the reaction in reactant A (what is the value of m)?
k1 [B]
0.0034 0.200 M
0.0020 0.100 M
0.0010 0.050 M
Options:
A. 0
B. 1
C. 2
D. 3
Answer:
Explanation:
Consider the given reaction,
A+B +C
[tex]rate=k[A]^m[B]^m[/tex]
Reaction is first order with respect to reactant B.
The pseudo order rate is
[tex]rate=k_1 [B]^m,\\\\where k_1 =k[A]^m[/tex]
Pseudo order reaction is the reaction in which the reactant concentration remains constant; its concentration is very high so it does not involve in the rate of the reaction. Here, the concentration of A is sufficiently high so this reaction becomes the pseudo first order reaction.
The reaction is first order with respect to [B] with rate constant k1 .
Using the given table, concentration of B is decreasing the rate also decreasing
Therefore, order of the reaction with respect to A is 1.
Value of m is 1.The order of the reaction in reactant A is 1, means the value of m for given rate should be 1.
What is pseudo order reaction?Pseudo order reactions are those reactions in which two reactants are present but the concentration of one of the reactant is present in excess quantity and it seems like first order reaction.
Given chemical reaction is: A + B +C
According to the question, given reaction is first order with respect to the reactant B and rate law will be represented as:
Rate = k[A]m[B]m
Also given that reactant A is present in high concentration in the reaction, so the rate of pseudo order reaction is:
Rate = k1[B]n, where
k1 = k[A]m
According to the given table with respect to concentration of reactant B and rate k1, it is clear that rate of the reaction decreases as the concentration of reactant B decreses. So, to get the required result order of the reaction or value of m with respect to the reactant A should be 1.
Hence option (B) is correct i.e. 1.
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A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128 J/g oF; water = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium
Answer: The final temperature of both the weight and the water at thermal equilibrium is [tex]50.26^{o}C[/tex].
Explanation:
The given data is as follows.
mass = 7.62 g, [tex]T_{2} = 10.8^{o}C[/tex]
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q = [tex]mC \times \Delta T[/tex]
= [tex]7.62 g \times 4.184 J/^{o}C \times (52.3 - T)[/tex]
Now, heat gained by lead will be calculated as follows.
q = [tex]mC \times \text{Temperature change of lead}[/tex]
= [tex]2.04 \times 0.128 \times (T - 11.0)[/tex]
According to the given situation,
Heat lost = Heat gained
[tex]7.62 g \times 4.184 J/^{o}C \times (52.3 - T)[/tex] = [tex]2.04 \times 0.128 \times (T - 11.0)[/tex]
T = [tex]50.26^{o}C[/tex]
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is [tex]50.26^{o}C[/tex].
The conversion of methyl isonitrile to acetonitrile in the gas phase at 250 °C CH3CN(g) is first order in CH3NC with a rate constant of 3.00×10^-3 s-1. If the initial concentration of CH3NC is 5.45×10^-2 M, the concentration of CH3NC will be 1.56×10^-2 M after_____________ s have passed.
Answer: The concentration of [tex]CH_3NC[/tex] will be [tex]1.56\times 10^{-2}M[/tex] after 416 seconds have passed.
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant = [tex]3.00\times 10^{-3}s^{-1}[/tex]
t = age of sample = ?
a = let initial amount of the reactant = [tex]5.45\times 10^{-2}M[/tex]
a - x = amount left after decay process = [tex]1.56\times 10^{-2}M[/tex]
[tex]t=\frac{2.303}{3.00\times 10^{-3}}\log\frac{5.45\times 10^{-2}}{1.56\times 10^{-2}}[/tex]
[tex]t=416s[/tex]
The concentration of [tex]CH_3NC[/tex] will be [tex]1.56\times 10^{-2}M[/tex] after 416 seconds have passed.
A 2.95 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder that contains 12.13 mL H2O. What will be the final volume reading on the graduated cylinder?
Answer:
12.54mL
Explanation:
Step 1:
Data obtained from the question:
Mass of Zn = 2.95g
Density of Zn = 7.14 g/mL
Volume of water = 12.13mL
Final volume reading of cylinder =..?
Step 2:
Determination of the volume of Zn. The volume of Zn can be obtained as follow:
Density = Mass /volume
7.14 = 2.95/volume
Cross multiply
7.14 x Volume = 2.95
Divide both side by 7.14
Volume = 2.95/7.14
Volume of Zn = 0.41mL
Step 3:
Determination of the final volume reading of the cylinder.
The final reading of the cylinder will be the addition of the volume of water and the volume of Zn as shown below:
Final volume of cylinder = volume of water + volume of Zn.
Volume of water = 12.13mL
Volume of Zn = 0.41mL
Final volume =..?
Final volume of cylinder = 12.13 + 0.41
Final volume of cylinder = 12.54mL
Therefore, the final volume reading of the cylinder is 12.54mL
The final volume reading of the graduated cylinder is 12.54 mL.
A 2.95 g piece of zinc (density = 7.14 g/mL) is added to a graduated cylinder. The volume of the piece of zinc is:
[tex]2.95 g \times \frac{1mL}{7.14 g} = 0.413 mL[/tex]
The final volume reading of the graduated cylinder will be equal to the sum of the volumes of the piece of zinc and the water.
[tex]V = V(H_2O)+V(Zn) = 12.13 mL + 0.413 mL = 12.54 mL[/tex]
The final volume reading of the graduated cylinder is 12.54 mL.
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if two atoms are bonded to a central atom with no lone pairs , how will they be arranged?
At a festival, spherical balloons with a radius of 170.cm are to be inflated with hot air and released. The air at the festival will have a temperature of 25°C and must be heated to 100°C to make the balloons float. 1.00kg of butane C4H10 fuel are available to be burned to heat the air. Calculate the maximum number of balloons that can be inflated with hot air.
Answer:
The correct answer is 1.5 balloons.
Explanation:
Based on the given question, the radius of the spherical balloons is 170 cm. Therefore, the volume of each balloon will be,
= 4/3 × π × (170)³ cm³
= 20.5176 × 10⁶ cm³
= 20.5176 m³
The density of air at 100 degree C s 0.946 Kg m⁻³
The mass of air in each balloon can be calculated by using the formula,
Mass = density × volume
Mass = 0.946 Kg m⁻³ × 20.5176 m³
Mass = 19.410 Kg
The heat energy, that is, required to bring the air from 25 degree C to 100 degree C will be,
= 19.410 × 10³ g × (100 -25) degree C × 1.009 J/g degree C
= 14.68 × 10⁵ J
The concentration of butane given is 1.00 Kg or 1000 g
The molecular weight of butane is 58.12 g per mole
The moles or n can be calculated by using the formula,
n = mass / mol.wt
n = 1000 g/58.12 g/mol = 17.20 mol
The formation of enthalpy of butane at 25 degree C is 125.7 × 10³ J/mol. The evolution of heat energy that take place at the time of burning 17.20 moles of butane is,
= 125.7 × 10³ J/mol × 17.20 mol
= 2.16 × 10⁶ J
The number of balloons that can be inflated with hot air is,
= 2.16 × 10⁶ J / 14.68 × 10⁵ J per each balloon
= 1.5 balloons
Hence, maximum of 1.5 balloons can be inflated.
Many computer chips are manufactured from silicon , which occurs in nature as SiO2. WhenSiO 2 is heated to melting , it reacts with solid carbon to form liquid silicon and carbon monoxide gas. In an industrial preparation of silicon, 155.8 kg of SiO2 reacts with 78.3 kg of carbon to produce 66.1 kg of silicon. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.
Answer:
In the given case, SiO2 is the limiting reactant, 71.72 Kg is the theoretical yield, and 90.8 percent is the experimental yield.
Explanation:
A reactant that completely gets used up in a chemical reaction, and it limits the concentration of product, which can be produced is termed as a limiting reactant. The maximum concentration of product that can get generated from the limiting reactant is termed as the theoretical yield.
Based on the given question, the balanced chemical reaction is:
SiO₂ + 2C ⇒ Si + 2CO
Now the mole ratio of SiO₂: C = 1: 2
Based on the given information, the mass of SiO₂ is 155.8 Kg. The no. of moles can be calculated by using the formula, n = weight/molecular mass.
The molecular mass of SiO₂ is 60.08 gram per mole. By putting the values we get,
n = 155.8 × 10³ grams / 60.08 grams/mol
n = 2.59 × 10³ moles
The weight of Carbon given is 78.3 Kg or 78.3 × 10³ grams, the molecular mass of C is 12 gram per mole. Similarly, the moles of C will be,
n = 78.3 × 10³ grams / 12 g/mol
n = 6.52 × 10³ moles
The experimental molar ratio of SiO₂: C is,
= 2.59 × 10³: 6.52 × 10³
= 2.59: 6.52 = 1: 2.5
Hence, it is clear that carbon in the given case is present in excess amount, therefore, SiO₂ is a limiting reactant.
With the help of the balanced equation, the molar ratio of SiO₂: Si is 1: 1. As SiO₂ is the limiting reactant, therefore, theoretical yield of the reaction will be determined by SiO₂.
As calculated, the moles of SiO₂ used is 2.59 × 10³
Hence, the moles of Si produced in the given reaction is 2.59 × 10³.
Now the mass of Si produced theoretically can be determined by using the formula, moles = weight/molar mass. The molar mass of Si is 28.08 gram per mole. Now putting the values we get,
Weight = 2.59 × 10³ moles × 28.08 gram per mole
Weight = 72.72 × 10³ grams or 72.72 Kg
The theoretical yield of Si is 72.72 Kg, however, the experimental yield is 66.1 Kg. Therefore, the percent yield for the reaction will be,
= Experimental yield / Theoretical yield × 100 %
= 66.1 Kg / 72.72 Kg × 100%
= 90.8 %
g 1. An ideal gas is confined to a container with adjustable volume. The number of moles, n, and temperature, T, are constant. By what factor will the volume change if pressure decreases by a factor of 8.5 Hint: pV
Answer:
The volume will increase by a factor of 8.5
Explanation:
In an ideal gas, the volume is inversely proportional to the pressure. This means that the volume will decrease when the pressure increases and vice versa.
Let the initial volume be v1, and the initial pressure p1.
the final pressure is decreased by a factor of 8.5, i.e
p2 = p1/8.5 = 0.12p1
using p1v1 = p2v2
v2 = (p1/p2)v1
v2 = (p1/0.12p1)v1
v2 = 8.5v1
This means that the final pressure increases by a factor of 8.5
Answer:
Increases by a factor of 8.5
Explanation:
Hello,
In this case, with the given data, we can apply the Boyle's law in order to understand the pressure-volume behavior as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, for the given pressure decrease we have:
[tex]P_2=\frac{P_1}{8.5}[/tex]
So the volume changes by:
[tex]V_2=\frac{P_1V_1}{\frac{P_1}{8.5} } \\\\V_2=8.5V_1[/tex]
Therefore, it increases by a factor of 8.5.
Regards.
Imagine you are volunteering in a 1st grade classroom. The students recently learned about the water cycle. One student raises her hand and asks, "I heard that elements like carbon can cycle too. How does carbon cycle? Is it like how water cycles?" How would you explain the cycling of carbon to this 1st grade class? Keep in mind that the students have likely not taken an advanced science course yet!
Answer:
Carbon cycle is also important for the survival of life on earth.
Explanation:
In carbon cycle, carbondioxide gas is present in the atmosphere. This carbondioxide gas is taken inside by the plant body through stomata and made food from it. When this food is eaten by animals, this carbon is transferred to their bodies. During breathing, carbondioxide gas is removed from animal body and the carbondioxide gas again goes to the atmosphere.
Consider an Al-4% Si alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at 578˚C, (d) the amounts and compositions of each phase at 576˚C, (e) the amounts and compositions of each microconstituent at 576˚C, and (f) the amounts and compositions of each phase at 25˚C.
Answer:
(a) Hypoeutectic
(b) Alpha solid, aluminium
(c) 70% α , 30% β
(d) 97.6% α, 2.4% β
(e) 97.6% α, 2.4% β
(f) 97% α, 3% β
Explanation:
(a) The eutectic composition for Al Si alloy is 11.7 wt% silicon, therefore, an Al-4% Si alloy is hypoeutectic
(b) For the hypoeutectic alloy, aluminium, Al, is expected to form first, such that the aluminium content is reduced till the point it gets to the eutectic proportion of 11.7 wt% silicon
(c) At 578°C we have
% α: Al (11 - 4)/(11 - 1) = 70% α
% L: Si 100 - 70 = 30% β
(d) At 576°C we have
α: 99.83% Si (99.83 - 4)/(99.83- 1.65) = 97.6% α
β: 1.65% Si (4 - 1.65)/(99.83- 1.65) = 2.4% β
(e) Primary α: 1.65% α (99.83 - 4)/(99.83 - 1.65) = 97.6% α
Eutectic 4% Si = 100 - 97.6 = 2.4% β
(f) At 25°C we have;
α%: (99.83 - 4)/(99.83 - 1) = 97% α
β%: 100 - 97 = 3% β.
Calculate the final temperature of 12.0 g of Argon (considered as an ideal gas) that is expanded reversibly and adiabatically from V1 = 1.00 dm3 at T1 = 273.15 K to V2 = 3.00 dm3 . CV,m(Ar) = (3/2) * R.
Answer:
Explanation:
For adiabatic change the expression is
= [tex]PV^\gamma=constant[/tex]
[tex]=(\frac{RT}{V})V^\gamma = constant[/tex]
[tex]=TV^{\gamma-1} = constant[/tex]
[tex]=T_1V_1^{\gamma-1} =[/tex][tex]=T_2V_2^{\gamma-1}[/tex]
for Argon γ = 1.67
273.15 x 1 = T₂ x [tex]3^{1.67-1}[/tex]
T₂ = 273.15 / [tex]3^{0.67}[/tex]
= 273.15/ 2.0877
= 130.83 K.
How sensitive to changes in water temperature are coral reefs? To find out, measure the growth of corals in aquariums where the water temperature is controlled at different levels. Growth is measured by weighing the coral before and after the experiment.
(a) What are the explanatory and response variables?
explanatory variable:
response variable:
(b) Are they quantitative or categorical?
The explanatory variable is:
The response variable is:
Answers:
(a) The explanatory variable is the change in the temperature .
The response variable is the coral survivance.
(b) The explanatory variable is quantitative.
The response variable is categorical.
Explanation:
A quantitative variable is those that can take a range of values. In this case, the temperature may vary across a range of values associated with the survival of coral reefs.
A categorical (also known as qualitative) variable is characterized to present finite states. In this case, the coral reefs can survive or not depending on their responses to changes in the temperature.
Response and explanatory variables are used in experimental procedure ; the explanatory and response variables are :
Measured variable - > Change in weight of coral (quantitative) Explanatory variable - > Temperature of water. (categorical)Explanatory variables are independent variables which causes changes to the measured variable.
Measured variable is the variable we intend to measure in an experiment. They are called dependent variables.
Quantitative variables are numerical values, weight is measured as a number. Hence, it is quantitative.
Categorical variables are labels and are used to group objects into classes. They are Non-numeric. Levels of temperature could be (high, medium, low). Hence, temperature is categorical in the experiment.
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describe how the relative density of a salt solution can be determined using the hydrometer
The hydrometer is an instrument used to determine the density of liquids, and looks like a large glass float filled at the bottom with grains of lead or mercury.
Because the mass of the hydrometer is known and thus the mass of the displaced liquid, the density can be determined using the volume.
There is a scale on the handle, the density can be read here. The deeper the hydrometer sinks, the lower its density.
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Eduardo hurts his knee during a basketball game. The trainer applies a cold pack, which gets cold after being squeezed. The next day, Eduardo’s friend Beth is going sledding. Beth buys some hand warmers and puts them into her gloves. All morning, her hands stay toasty warm.
Required:
a. How do you think these devices work?
b. Where do you think the "cold" and the heat comes from?
Answer:
How do you think these devices work?
Both devices follows the principle and endothermic and exothermic reactions.The cold pack is made up of two bags. The first bag contains water.And it is placed inside the second bag which contains NH4NO3 , and urea. Therefore, when the water bag is squeezed, the bag busted releasing the water which dissolves the surrounding solid compounds. These compounds absolved heat energy in an endothermic reaction.
The components of hand warmers are activated carbon, salt iron, water,,and cellulose.The Oxidation of iron when exposed to the air (oxygen)generated heat in an exothermic reaction to keep the hand warm.
Thus the cold results from the endothermic reaction as the solid compounds in cold packs absorbed water to break bonds to melts,and the hotness comes from formation of new bonds(exothermic) as oxidation of Fe to form new compounds with other chemical occurs.
Explanation:
PLEASE HELP I HAVE LIMITED TIME!!
How many particles of water (H20) are in a collection of snowflakes
with a mass of 0.005 g?
Answer:
Explanation:
Si tomamos en cuenta el peso molecular del agua, que es equivalente a:
1 Átomo de H₂O
O = 16 gr/mol
H = 1 gr/mol
H₂O = 18 gr/mol
Teóricamente sabemos que en 1 mol de H2O habrá 18 gr.
Para obtener los moles presentes en 1 mg de H₂O, (como 1 gr = 1000 mg), decimos:
1 mol H2O ………………………….. 18000 mg
X …………………………… 1 mg
X = 1 / 18000 = 5,56 X 10⁻⁵ moles de H20
Y para obtener la cantidad de moléculas presentes, de acuerdo a los moles, multiplicamos por el número de Avogadro (6,023 X 10²³ moléculas /mol)
Moléculas de H₂O = 5,56x 10⁻⁵ mol x 6,023 x 10²³
Moléculas de H₂O = 3,34488 x10¹⁹ moléculas de H₂O
En el copo de nieve habrá 3,34488 x 10¹⁹ moléculas de H₂O.
Espero que te sirva =)
5,43 X 10²² particles of water (H20) are in a collection of snowflakes with a mass of 0.005 g.
What do you mean by mole ?The term mole is defined as the amount of substance of a system which contains as many elementary entities.
One mole of any substance is equal to 6.023 × 10²³ units of that substance such as atoms, molecules, or ions. The number 6.023 × 10²³ is called as Avogadro's number or Avogadro's constant.
The mole concept can be used to convert between mass and number of particles.
1 atom of H₂O
O = 16 g/mol
H = 1 gr/mol
H₂O = 18 gr/mol
1 mol H₂O ………………………….. 18000 mg
? X …………………………… 1 mg
X = 1 / 18000
= 5,43 X 10²² moles de H20
Thus, 5,43 X 10²² particles of water (H20) are in a collection of snowflakes with a mass of 0.005 g.
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3. The exosphere is the outer layer of the
O D. thermosphere
OC. mesosphere
O A. ozone layer
O B. troposhpere
ang
AT
HHH
Answer:
D. exosphere is the outer layer of the thermosphere
Sort The following iron compound by whether the cation is iron(II) or iron(III)
Answer:
Left, Left, Right, Left, Right, Right
Explanation:
Follow that order!
Sodium Chloride (NaCl) has a molecular mass of 58 grams/mole. What is the molarity of a NaCl solution with a
concentration of 232 grams/liter?
2.0 molar
4.0 molar
0.5 molar
3.0 molar
Answer:
2nd option
Explanation:
Molarity is the number of moles of the solute (NaCl) in 1 litre of the solution (NaCl solution).
Given: concentration= 232g/ L
what we are trying to achieve is __mol/ L.
So in 1 litre, we have 232g of NaCl.
To convert mass to mole, we divide it by the Mr.
Given that the Mr is 58g/mol,
number of moles
= 232 ÷58
= 4
Thus, 1 litre has 4 moles of NaCl.
Therefore, the molarity is 4.0 mol/L.
The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 760.0 torr and the vapor pressure of water under the experimental conditions is 23.8 torr, what is the pressure (torr) of the H 2 gas
Answer:
the pressure (torr) of theH₂ gas is 736.2 torr
Explanation:
Given that
The H₂ gas produced in a chemical reaction is collected through water in a eudiometer; during this process, the gas collected contains some droplets of water vapor along with these gas.
So; the total pressure in the eudiometer = Pressure in the H₂ gas - Pressure of the water vapor
Where;
[tex]P_{Totsl}[/tex] = total pressure in the eudiometer = 760.0 torr
[tex]P_{H_2}[/tex] = Pressure in the H₂ gas = ???
[tex]P_{H_2O}[/tex] = Pressure in the water vapor = 23.8 torr
Now:
[tex]P_{Totsl}[/tex] = [tex]P_{H_2}[/tex] + [tex]P_{H_2O}[/tex]
- [tex]P_{H_2}[/tex] = + [tex]P_{H_2O}[/tex] - [tex]P_{Totsl}[/tex]
[tex]P_{H_2}[/tex] = - [tex]P_{H_2O}[/tex] + [tex]P_{Totsl}[/tex]
[tex]P_{H_2}[/tex] = (- 23.8 + 760) torr
[tex]P_{H_2}[/tex] = 736.2 torr
Thus; the pressure (torr) of theH₂ gas is 736.2 torr
If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.
A eudiometer is a laboratory device that measures the change in volume of a gas mixture following a physical or chemical change. Usually, the gas is collected over water.
According to Dalton's law of partial pressures, the sum of the sum partial pressure of the hydrogen and the partial pressure of the water is equal to the total pressure in the eudiometer.
[tex]P = pH_2 + pH_2O\\\\pH_2 = P - pH_2O = 760.0 torr - 23.8 torr = 736.2 torr[/tex]
If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.
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The bitter taste often associated with leafy green vegetables such as spinach, parsley, and cilantro is often due to oxalic acid, which has a molecular weight of 90.04g. The percentage of the acid present can be determined through an acid-base titration with NaOH once the sample has been prepared and the oxalic acid is extracted into an aqueous solution. A 100.0-g sample was treated so that all of the oxalic acid dissolved. Because the solution was intensely green due to chlorophyll, an indicator could not be used, and the titration was monitored using pH meter. The following data were obtained. Concentration of NaOH -0.693 M volume 0.00 pH 1.63 2.02 14.00 1.01 12.00 2.49 2.46 10.00 4.56 2.80 8.00 6.01 2.98 pH 7.69 3.16 6.00 8.33 3.23 4.00 3.51 10.72 13.35 2.00 3.91 14.14 4.09 0.00 0 5 10 1516 20 15.41 4.67 15.67 4.97 Volume of NaOH 15.85 5.49 5.79 15.89 (a) How many moles of oxalic acid were present in the 100- g sample? 15.91 6.09 15.95 12.14 16.01 12.74 16.52 13.61 17.06 13.65 (b) Calculate the percentage of oxalic acid in the spinach.
Answer:
1. Moles of oxalic acid in 100 g sample = 0.0055moles
2. Percentage oxalic acid in 100g sample = 0.5%
Explanation:
Check attachment below for explanation and correct chart arrangement
which statement best describle for formation of solution
Explanation:
small amount of solute is dissolved in a large amount of solvent. it would not make sense to have more solute and solvent. you can think of it like this, you can dissolve a teaspoon of salt in a litre water but you can't resolve 1 kg of salt in a teaspoon of water.
also, the solute is what is begin dissolved and the solvent is what dissolving the solute so the one day the solvent is dissolved in the solute does not make complete sense.
A salt contains a compound containing a metal and a nonmetal. This is an
compound. *
A.molecular
B.covalent
C.Tonic
D.None of the above
Answer:
Option C. Ionic
Explanation:
Ionic bond is formed between a metal and a non metal.
Ionic bond occurs when there is a transfer of electron(s) from a metallic atom to a non metallic atom.
What is the molarity of a solution of 3 moles (mol) of FeBr3 in 1/2 L water?
Answer:
3mol/0.5liters or 6 M
Explanation:
The molarity of a solution of 3 moles (mol) of FeBr₃ in 1/2 L water is 6 Molar Solution.
What is Molarity ?Molarity (M) is the amount of a substance in a certain volume of solution.
Molarity is defined as the number of moles of a solute per liters of a solution.
Molarity is also known as the molar concentration of a solution.
Molarity is number of moles of solute in 1 litre of solution.
Therefore,
If 0.5 ltr of solution consisting of 3 moles of FeBr₃
Thus,
1 ltr of solution consisting of 6 moles of FeBr₃.
6 moles of solute in 1 ltr of solution means 6 Molar solution.
Hence, The molarity of a given solution of 3 moles (mol) of FeBr₃ in 1/2 L water is 6 Molar Solution.
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What is mass per unit volume called?
Answer:
The correct answer is Density
Explanation:
Hope this helps you
Answer:density
Explanation:
One kilogram is the same as
Answer:
1000 grams
Explanation:
Kilo- is the SI prefix meaning 1000. So, 1 kilogram is the same as 1000 grams.
How many hydrogen atom are present in 1.53 g of water
Answer:
[tex]=1.02x10^{23} atoms\ H[/tex]
Explanation:
Hello,
In this case, for water, whose molar mass is 18 g/mol, we can find two moles of hydrogen in one mole of water, therefore, for us compute the atoms, we should also use the Avogadro's number as shown below:
[tex]=1.53gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} *\frac{6.022x10^{23}atoms\ H}{1molH} \\\\=1.02x10^{23} atoms\ H[/tex]
Regards.
