Answer:
Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.
Explanation:
To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask
The molar mass of Aluminum Sulfate = 342.15 g/mol
Using the molarity formula:-
Molarity = Number of moles/Volume of solution in a liter
Number of moles = Given weight/ molar mass
Molarity = (Given weight/ molar mass)/Volume of solution in liter
0.187 M = (Given weight/342.15 g/mol)/0.250 L
Given weight = 15.99 g
Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O. Suppose 55.8 g of hydrobromic acid is mixed with 17. g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Round your answer to 2 significant digits.
Answer:
21.4g of HBr is the minimum mass that could be left over.
Explanation:
Based on the reaction:
HBr + NaOH → NaBr + H2O
1 mole of HBr reacts per mole of NaOH
To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:
Moles NaOH -40.0g/mol-
17g * (1mol/40.0g) = 0.425 moles NaOH
Moles HBr -Molar mass: 80.91g/mol-
55.8g * (1mol/80.91g) = 0.690 moles HBr
The difference in moles is:
0.690 moles - 0.425 moles =
0.265 moles of HBr could be left over
The mass is:
0.265 moles * (80.91g/mol) =
21.4g of HBr is the minimum mass that could be left over.Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion: Fe3+(aq) + 3 C2O42-(aq) ⇌ [Fe(C2O4)3]3-(aq). If 0.0800 M Fe3+ is initially mixed with 1.00 M oxalate ion, what is the concentration of Fe3+ ion at equilibrium?
Answer:
hello? are you still here? reply if you are
tính ΔH° của phả ứng sau ở 200°C
CO+1÷2O=CO2
ΔH°
Write the balanced equation for the hydration of CuSO4CuSO4. Indicate the physical states using the abbreviations (ss), (ll), or (gg) for solid, liquid, or gas, respectively. Use (aqaq) to indicate the aqueous phase. Indicate appropriate charges on negative and positive ions if they are formed.
Answer:
CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)
Explanation:
Hydration is the process by which anhydrous CuSO4 acquires molecules of water of crystalization to form the pentahydrate.
The water of crystalization becomes attached go the crystals of the CuSO4 to form the hydrated salt.
Beginning with solid anhydrous CuSO4 we have;
CuSO4(s) + 5H2O(l) ----> CuSO4.5H2O(s)
Radon-220 undergoes alpha decay with a half-life of 55.6 s.?
Assume there are 16,000 atoms present initially and calculate how many atoms will be present at 0 s, 55.6 s, 111.2 s, 166.8 s, 222.4 s, and 278.0 s (all multiples of the half-life). Express your answers as integers separated by commas.
Calculate how many atoms are present at 50 s, 100 s, and 200 s (not multiples of the half-life).
The half life of a radioactive isotope refers to the time taken for half of the number of original number of atoms present in the sample to decay.
The equation below gives the number of atoms present at time t
[tex]N=Noe^-kt[/tex]
N = Number of atoms present at time t
No = Number of atoms initially present
k = decay constant
t = time taken
Given that;
t1/2 = 0.693/k
where t1/2 = half life
k = 0.693/t1/2
k = 0.693/ 55.6 s
k = 0.0125 s-1
Substituting values;
N = 16,000 e^-0.0125(0)
N = 16,000 atoms
At 50 s
N = 16,000 e^-0.0125(50)
= 8564 atoms
At 100 s
N = 16,000 e^-0.0125(100)
= 4584 atoms
At 200 s
N = 16,000 e^-0.0125(200)
= 1313 atoms
https://brainly.com/question/2998270
What particules make up the nucleus
Answer:
nucleus is a collection of particles called protons,which are positively charged..and neutrons which are electrically neutral..electrons which are negatively charged..and neutrons are in turn made up of particles called quarks ..
Explanation:
hope this helps u ...
Answer:
The Nucleus is made up of protons and neutrons.
The mole fraction of NaCl in an
aqueous solution is 0.132. How
many moles of NaCl are present in
1 mole of this solution?
Molar Mass
NaCl: 58.44 g/mol
H2O: 18.016 g/mol
Answer:
Moles of water are 0.868
Explanation:
An analytical chemist is titrating of a solution of benzoic acid with a solution of . The of benzoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added.
The question is incomplete. The complete question is :
An analytical chemist is titrating 148.9 mL of a 1.100 M solution of benzoic acid [tex]$HC_6H_5CO_2$[/tex] with a 0.3600 M solution of KOH. The [tex]pK_a[/tex] of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 232.0 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added.
Solution :
Number of moles of [tex]$C_6H_5OCOOH$[/tex] [tex]$=148.9 \ mL \times \frac{L}{1000\ mL} \times \frac{1.100 \ mol}{L}$[/tex]
= 0.16379 mol
Number of moles of NaOH added [tex]$=232.0 \ mL \times \frac{L}{1000\ mL} \times \frac{0.3600 \ mol}{L}$[/tex]
= 0.08352 mol
ICE table :
[tex]C_6H_5OCOOH \ \ \ + \ \ \ OH^- \ \ \ \rightarrow \ \ C_6H_5OCOO^- \ \ \ \ + \ \ H_2O[/tex]
I (mol) 0.16379 0.08352 0
C (mol) -0.08352 -0.08352 +0.08352
E (mol) 0.08027 0 0.08352
Total volume = (148.9 + 232) mL
= 380.9 mL
= 0.3809 L
Concentration of [tex]$C_6H_5OCOOH, [C_6H_5OCOOH]$[/tex] [tex]$=\frac{0.08027 \ mol}{0.3809 \ L}$[/tex]
= 0.211 M
Concentration of [tex]$C_6H_5OCOO^- , [C_6H_5OCOO^-] =\frac{0.08352 \ mol}{0.3809 \ L}[/tex]
= 0.219 M
[tex]pK_a[/tex] of [tex]C_6H_5OCOOH = 4.20[/tex]
According to Henderson equation,
[tex]$pH = pK_a + \log \frac{[C_6H_5OCOO^-]}{[C_6H_5OCOOH]}[/tex]
[tex]$=4.20 + \log \frac{0.219}{0.211}$[/tex]
= 4.22
Therefore, the pH of the acid solution is 4.22
A person uses 400.8 kcal of energy to run a race. Convert the energy used for the race to the following energy units:
(provide an answer in 4 significant figures)
Calories
calories
Food Calories
Joules
Kilojoules
Hint: 1kcal=4.184kJ
Calories, we know that fat burn is calories.
Different steps of the oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase PDH complex are given. Place these five steps in the correct order. Note that thiamine pyrophosphate, TPP, is sometimes called thiamine diphosphate, TDP.
1. FADH2 is reoxidized to FAD reducing NAD* to NADH.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
Answer:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
Explanation:
The oxidation of pyruvate to AcetylCoA is catalyzed by the pyruvate dehydrogenase complex. The reaction is an irreversible oxidative decarboxylation process in which the carboxyl group of pyruvate is removed as a molecule of carbon dioxide, CO₂, while the remaining two carbons are attached to a CoASH molecule to form acetylCoA.
The pyruvate dehydrogenase complex contains three enzymes - Pyruvate dehydrogenase known as E₁, dihydrolipoyl transacetylase known as E₂, and dihydrolipoyl dehydrogenase known as E₃. It also requires five coenzymes namely: thiamine pyrophosphate (TPP), flavine adenine dinucleotide (NAD), coenzyme A (CoA-SH), nicotinamide adenine dinucleotide (NAD) and lipoate.
Oxidative decorbyxylation of pyruvate takes place in the pyruvate dehydrogenase complex in five steps:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
Write a chemical equation for LiOH(aq) showing how it is an acid or a base according to the Arrhenius definition.
Answer:
LiOH(aq) → Li⁺(aq) + OH⁻(aq).
A bond dissociation energy is A) The energy required to remove an electron from an atom. B) The energy released when an ionic compound dissociates in water. C) The energy required to break a covalent bond. D) The energy produced in a chemical reaction that breaks chemical bonds.
Answer:
The energy required to break a covalent bond
Explanation:
When a chemical bond is formed, energy is released. When a chemical bond is broken, energy is absorbed.
We define the bond dissociation energy as the energy required to break a covalent bond. The process of covalent bond cleavage is endothermic hence energy is absorbed for the process to occur.
Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g)2FeO(s)...... ΔH° = -544.0 kJ (2) 2Zn(s) + O2(g)2ZnO(s)......ΔH° = -696.6 kJ what is the standard enthalpy change for the reaction:
Answer:
-76.3 kJ
Explanation:
Here is the complete question
Given the standard enthalpy changes for the following two reactions:
(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ
(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:
(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?
Solution
Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ
reversing the reaction, we have
2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)
Adding reactions (2) and (3), we have
2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)
2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ (2)
This gives
2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =
The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)
= +544.0 kJ - 696.6 kJ)
= -152.6 kJ
Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)
we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).
So, ΔH° = -152.6 kJ/2 = -76.3 kJ
So, the standard enthalpy change for the reaction
FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ
Please guys please answer this
Answer:
1. Because the rules will keep you safe it prevents you from getting hurt.
2.i) don't taste chemical
ii) Always wear protective gears
iii) be careful with tool
iv) wear protective gloves
Explanation:
help everyone get out quickly
What is the balanced equation for the reaction of lithium metal with fluorine gas? Li ( s ) + F ( g ) → LiF ( s ) Li ( s ) + F 2 ( g ) → LiF ( s ) 2 Li ( s ) + F 2 ( g ) → 2 LiF ( s ) Li ( s ) + F 2 ( g ) → LiF
Answer:
2Li(s) + F2(g)→2LiF(s)
Explanation:
The radioactivity due to carbon-14 measured in a piece of a wood from an ancient site was found to produce 20 counts per minute from a given sample, whereas the same amount of carbon from a piece of living wood produced 160 counts per minute. The half-life of carbon-14, a beta emitter, is 5730 y. The age of the artifact is closest to
Answer:
The answer is "17200 years".
Explanation:
Given:
[tex]A = 20 \ \frac{counts}{minute}\\\\A_{o} = 160\ \frac{counts}{minute}[/tex]
Let the half-life of carbon-14, is beta emitter, is [tex]T = 5730\ years[/tex]
Constant decay [tex]\ w = \frac{0.693}{ T}[/tex]
[tex]= 1.209 \times 10^{-4} \ \frac{1}{year}\\[/tex]
The artifact age [tex]t= ?[/tex]
[tex]A = A_{o} e^{-wt} \\\\e^{-wt} = \frac{A}{A_{o}}\\\\-wt = \ln \frac{A}{A_{o}}\\\\= -2.079\\\\t = 1.7199 \times 10^{4} \ years\\\\\sim \ 17200\ years\\[/tex]
The idea that the behavior of the states of matter is determined by the kinetic energy and movement of their particles is called _____…
A. Sublimation Theory
B. Kinetic Movement Theory
C. Kinetic Molecular Theory
D. Van der Waals Theory
Answer:
C . Kinetic Molecular Theory
list two uses of H2SO4
Based on the reaction below:
[tex]N_2 + 3H_2[/tex] ↔ [tex]2NH_3 + heat[/tex]
If we decrease the temperature, equilibrium will shift towards the...
Please explain!
N₂ + 3H₂ ⇄ 2NH₃ + heat
In the given equilibrium, we notice that the heat is on the right. which means that if the heat requirements don't meet, the reactants on the right will no longer react due to the lack of heat
but because the reactants on the left don't have such weaknesses, they will keep reacting hence producing more and more ammonia until a new equilibrium is reached
where there will be more ammonia and less nitrogen and hydrogen as compared to the equilibrium we had initially
Answer:
Explanation:
heat is given out as 1 of the products, along w/ NH3 in the forward reaction. so its an exothermic reaction
decreasing temperature favors exothermic reaction as more heat can be absorbed by the environment
so equilibrium will shift towards the products
Determine the total pressure of a mixture that contains 5.25 g of He and 3.25 g of N2 in a 7.75-L flask at a temperature of 27ºC.
Answer:
4.54 atm
Explanation:
Step 1: Calculate the total number of gaseous moles
We will calculate the moles of each gas using its molar mass.
He: 5.25 g × 1 mol/4.00 g = 1.31 mol
N₂: 3.25 g × 1 mol/28.01 g = 0.116 mol
The total number of moles is:
n = 1.31 mol + 0.116 mol = 1.43 mol
Step 2: Convert 27 °C to Kelvin
We will use the following expression.
K = °C + 273.15 = 27 + 273.15 = 300 K
Step 3: Calculate the total pressure of the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 1.43 mol × (0.0821 atm.L/mol.K) × 300 K / 7.75 L = 4.54 atm
The solubility of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10â2 M.
a. Write the balanced solubility equilibrium equation for LiF.
b. Determine the molar concentration of the lithium ion and the fluoride ion.
c. Write the Ksp expression for the reaction.
d. Calculate Ksp for lithium fluoride.
Answer:
a. LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
b. [Li⁺] = [F⁻] = 6.2 x 10⁻² M
c. Ksp = [Li⁺] [F⁻]
d. Ksp = 3.8 × 10⁻³
Explanation:
The solubility (S) of lithium fluoride, LiF, is 1.6 g/L, or 6.2 x 10⁻² M.
a. The balanced solubility equilibrium equation for LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
b. We will make an ICE chart.
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
I 0 0
C +S +S
E S S
Then, [Li⁺] = [F⁻] = S = 6.2 x 10⁻² M
c. The solubility product constant, Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution.
Ksp = [Li⁺] [F⁻]
d.
Ksp = [Li⁺] [F⁻] = (6.2 x 10⁻²)² = 3.8 × 10⁻³
A person slips over banana pills. Give reason
Answer:
We slip when we step on a banana peel because the inner side of banana peel being smooth and slippery reduces the friction between the sole of our shoe and the surface of road.
Net ionic reaction of H2SO4 with Ba(OH)2
Answer:
This is an acid-base reaction (neutralization): Ba(OH) 2 is a base, H 2SO 4 is an acid. This is a precipitation reaction: BaSO 4 is the formed precipitate.
write down the different uses of water that you know about
Answer:
The various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water is used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
Answer:
The various use of water are;
I) Cooking.
ii) Drinking
III) Bathing
iv) Generating hydro- electricity
v) Construction work etc
When hydrogen gas reacts with oxygen gas, water vapour is formed according to the
reaction 2H2 + O2 2H2O. If 3.00 mol of hydrogen gas react with 3.00 mol
of oxygen gas, which reactant will be the reactant in excess?
Explanation:
here's the answer to the question
The mass of an empty flask plus stopper is 64.232g. When the flask is completely filled with water the new mass is 153.617 g. The flask is emptied and dried, and a piece of metal is added. The mass of the flask, stopper and metal is 143.557. Next, water is added to the flask containing the metal and the mass is found to be 226.196. What is the density of the metal
Answer:
11.76 g/cm^3
Explanation:
Mass of empty flask and stopper = 64.232g
Mass of flask filled with water = 153.617 g
Mass of water = 153.617 g - 64.232g = 89.385 g
Mass of flask, stopper and metal = 143.557 g
Mass of metal = 143.557 g - 64.232g = 79.325 g
Mass of water, flask, stopper and metal = 226.196 g
Mass of water = 226.196 g - 143.557 g = 82.639 g
Since mass of water =volume of water
Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3
Density of metal = mass/volume = 79.325 g/6.746 cm^3
= 11.76 g/cm^3
what is the difference between 25ml and 25.00ml
Answer:
There is no difference between the two.
Explanation:
They both show the same volume. But, adding decimal places shows the least count of the instrument used and is more acceptable when recording values in scientific experiments
If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH can be added before the buffer capacity is reached
Answer:
25.5mL of 0.100M NaOH are needed to reach buffer capacity.
Explanation:
The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:
Moles sodium acetate / Moles Acetic acid = 10
The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:
HA + NaOH → H2O + NaA
That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.
The initial moles of each species is:
Acetic acid:
23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid
Sodium Acetate:
33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate
We can write the moles of each species when NaOH is added as:
Moles sodium acetate / Moles Acetic acid = 10
0.00623 moles + X / 0.00343 moles - X = 10
Where X are moles of NaOH added
Solving for X:
0.00623 moles + X = 0.0343 moles - 10X
11X = 0.0281
X = 0.00255 moles of NaOH are needed
In Liters:
0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =
25.5mL of 0.100M NaOH are needed to reach buffer capacity
The decomposition of ammonia is: 2 NH3(g) ⇌ N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.20 atm and H2 is 0.15 atm?
Answer:
"[tex]6.7\times 10^{-4} \ atm[/tex]" is the right answer.
Explanation:
Given:
Partial pressure of [tex]N_2[/tex],
= 0.20 atm
Partial pressure of [tex]H_2[/tex],
= 0.15 atm
[tex]K_p = 1.5\times 10^3[/tex] at [tex]400^{\circ} C[/tex]
As we know,
⇒ [tex]K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}[/tex]
By putting the values, we get
[tex]1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}[/tex]
[tex]pNH_3^2 = \frac{0.000675}{1.5\times 10^3}[/tex]
[tex]=6.7\times 10^{-4} \ atm[/tex]
melting points of lipids are strongly influenced by the length and degree of unsaturation of hydrocarbon chain.justify this statement?
Answer
The properties of fatty acids and of lipids derived from them are markedly dependent on chain length and degree of saturation. Unsaturated fatty acids have lower melting points than saturated fatty acids of the same length. For example, the melting point of stearic acid is 69.6°C, whereas that of oleic acid (which contains one cis double bond) is 13.4°C. The melting points of polyunsaturated fatty acids of the C18 series are even lower. Chain length also affects the melting point, as illustrated by the fact that the melting temperature of palmitic acid (C16) is 6.5 degrees lower than that of stearic acid (C18). Thus, short chain length and unsaturation enhance the fluidity of fatty acids and of their derivatives.
