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Likurg_2 [28]
4 years ago
11

Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. a

11.20-g mixture of zinc and magnesium produces 0.6854 g of hydrogen gas upon being mixed with an excess of hydrochloric acid. determine the percent magnesium by mass in the original mixture.
Chemistry
1 answer:
kirill115 [55]4 years ago
5 0
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?

m(Mg)=wm
m(Zn)=(1-w)m

Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)

Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)

m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}

(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}

w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)

w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))

w=0.583 (58.3%)
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Three 5-l flasks, fixed with pressure gauges and small valves, each contains 4 g of gas at 273 k. flask a contains h2, flask b c
Varvara68 [4.7K]
First, please check the missing part in your question in the attachment.
a) So first, the Rank of pressure:
according to this formula PV = nRT and when n = m/Mw
PV = m/Mw * R*T
when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater 
when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g
∴ Pressure :
 (A) > (B) > (c)

B) The rank of average molecular kinetic energy:
when K = 3/2 KB T
when K is the average kinetic energy per molecule of gas 
and KB is Boltzmann's constant
and T is the temperature (K)
So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:
∴ A = B = C
C) the rank of diffusion rate after the valve is opened:
according to this formula:
R2/R1 = √M1/M2
from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,
when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g
∴ the rank of diffusion:
A > B > C

D) The rank of the Total kinetic energy of the molecules:
when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules 
∵ Mw A < Mw B < Mw C 
∴no .of molecules of A > B >C
∴ the rank of total kinetic energy is:
A > B > C

e) the rank of density:

when ρ = m/ v 
and m is the mass & v is the volume and we have both is the same for A, B, and C
so the density also will be the same, ∴ the rank of the density is:
A = B = C

F) the rank of the collision frequency:
as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.
∴ Collision frequency will only depend on the no.of molecules
we have no.of molecules of A > B > C as Mw A < B < C 
∴the rank of the collision frequency is:
A > B > C 

 



7 0
3 years ago
What is the density of a substance that weighs 340g and has a volume of 40cm3?
jeka94

Answer:

<h2>8.5 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{340}{40}  =  \frac{17}{2}  \\

We have the final answer as

<h3>8.5 g/cm³</h3>

Hope this helps you

5 0
3 years ago
A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif
noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = [A_o]

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = [A]=0.06[A_o]

Half life of the carbon-14 = t_{1/2}=5730 years

k=\frac{0.693}{t_{1/2}}

[A]=[A_o]\times e^{-kt}

[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}

0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0
4 years ago
What effect does a quadrupling of the mass have upon the acceleration of the object
netineya [11]
The state in which all of the external forces acting upon an object are balanced; there is no acceleration. friction ..... quadrupling. doubling distance and quadrupling mass has the overall effect of the force
8 0
3 years ago
Determine the mass in grams of Avogadro's number of C12H22O11
Allushta [10]

Answer:

2.059524x10^26 if im not wrong

Explanation:

avogadro's number is 6.022x10^23

5 0
3 years ago
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