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ololo11 [35]
2 years ago
13

A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical formula? The first step is to calcula

te the mass of each element in a 100-g sample of the compound. We use 100 g because it makes the calculations more straightforward. What is the mass of each element in a 100.0-g sample of the compound? mass H = g mass S = g mass O = g
Chemistry
2 answers:
Georgia [21]2 years ago
7 0
For a 100-g compound, we would have 2 g H, 32.7 g S, and 65.3 g O. We then divide each of these masses by their molar masses:
2 g H / 1.01 g/mol = 1.98 mol H
32.7 g S / 32.07 g/mol = 1.0196 mol S
65.3 g O / 16.00 g/mol = 4.08125 mol O
We then divide each amount of moles by the smallest of them, which is the 1.0196 from S:
1.98 mol H / 1.0196 mol S = 1.94 H ~ 2 H
1.0196 mol S / 1.0196 mol S = 1 S
4.08125 mol O / 1.0196 mol S = 4 O
So the empirical formula of the compound is H2SO4.
anygoal [31]2 years ago
4 0

Answer:

2.00

32.7

65.3

in order

Explanation:

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One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori
Igoryamba

Answer:

P_2=0.398bar=39800Pa

T_2=118.7K\\

Q=-3729.9J

W=-61753.24J

ΔU_T=0J

ΔH_T=0J

Explanation:

Hello,

At the first state, the molar volume is:

v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3

The volume in both the second and third state:

v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3

Now, as it is about an adiabatic process, one remembers the following relationships:

PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa

- And the temperature:

T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J

Then the total heat:

Q=Q_1+Q_2=0-3729.9J=-3729.9J

- The work for the first process is:

W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

W=W_1+W_2=-61753.24J+0J=-61753.24J

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\

Best regards.

8 0
2 years ago
At one point in the video, someone is balancing a chemical reaction. The person writes a coefficient of 2 in front of Fe2O3 but
mario62 [17]

answer: the 4 is the product and the other numbers are reactants

Explanation:

i’m pretty sure that’s the answer . i did this last week :)

8 0
2 years ago
If the speed of a vehicle increases by 22%, by what factor does its kinetic energy increase?
Mila [183]

Answer:

The kinetic increases by 48.84 %

Explanation:

The expression for the kinetic energy is:-

K.E.=\frac{1}{2}\times mv^2

Where, m is the mass of the object

v is the velocity of the object

Let the new velocity is:- v'

v is increased by 22 %. Thus, v' = 1.22 v

So, the new kinetic energy is:-

K.E.'=\frac{1}{2}\times mv'^2=\frac{1}{2}\times m{1.22}^2=\frac{1}{2}\times mv^21.4884=1.4884K.E.

<u>Thus, the kinetic increases by 48.84 %</u>

7 0
2 years ago
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What volume would 0.435 moles of hydrogen gas, h2, occupy at stp?
faust18 [17]
The conversion factor for volume at STP is \frac {1mol}{22.4L} or \frac {22.4L}{1mol}. Since we want volume, we would use \frac {22.4L}{1mol}. We conclude with the following calculations:

0.435mol H_{2} * \frac {22.4LH_{2}}{1molH_{2}} = 9.744L

The answer is 9.744L H2
3 0
2 years ago
Read 2 more answers
Rusting of iron is a very common chemical reaction. It results in one form from Fe reacting with oxygen gas to produce iron (III
Vlada [557]

<u>Answer:</u> The given amount of iron reacts with 9.0 moles of O_2 and produce 6.0 moles of Fe_2O_3

<u>Explanation:</u>

We are given:

Moles of iron = 12.0 moles

The chemical equation for the rusting of iron follows:

4Fe+3O_2\rightarrow 2Fe_2O_3

  • <u>For oxygen gas:</u>

By Stoichiometry of the reaction:

4 moles of iron reacts with 3 moles of oxygen gas

So, 12.0 moles of iron will react with = \frac{3}{4}\times 12.0=9.0mol of oxygen gas

  • <u>For iron (III) oxide:</u>

By Stoichiometry of the reaction:

4 moles of iron produces 2 moles of iron (III) oxide

So, 12.0 moles of iron will produce = \frac{2}{4}\times 12.0=6.0mol of iron (III) oxide

Hence, the given amount of iron reacts with 9.0 moles of O_2 and produce 6.0 moles of Fe_2O_3

5 0
2 years ago
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