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-Dominant- [34]
3 years ago
5

How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26ÁC? Show all work used to solve this pr

oblem
Chemistry
1 answer:
klio [65]3 years ago
6 0
The answer is 7.33 g.

<span>To calculate this, we will use the the ideal gas law:
PV = nRT
where
P - pressure of the gas,
V - volume of the gas,
n - amount of substance of gas,
R - gas constant,
T - temperature of the gas.</span>

Since the amount of substance of gas (n) can be expressed as mass (m) divided by molar mass (M), then:

PV = RTm/M

It is given:

P = 0.98 atm

V = 10.2 l

T = 26°C = 299.15 K 

R = 0.082 l atm/Kmol (gas constant)

M (H2O) = 2Ar(H) + Ar(O) = 2*1 + 16 = 2 + 16 = 18g

m = ?

Since PV = RTm/M, then:

m = PVM/RT

m = 0.98 · 10.2 · 18 / 0.082 · 299.15 = 179.928/24.5303 = 7.33 g

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How many moles in 6.57 x 10^24 formula units of NaCl?
laiz [17]

Answer:

3.955*10^48

Explanation:

1 mole of a substance gives 6.02*10^23/6.57*10^24 will give x then cross multiply the answer. is 3.955*10^48

8 0
3 years ago
Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.
bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr <em>(1)</em>

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

5 0
3 years ago
A solution prepared by mixing 10 ml of 1 m hcl and 10 ml of 1.2 m naoh has a ph of
blagie [28]

Answer: pH of resulting solution will be 13

Explanation:

pH is the measure of acidity or alkalinity of a solution.

Moles of H^+ ion = Molarity\times {\text {Volume in L}}=1M\times 0.01L=0.01mol

Moles of OH^- ion = Molarity\times {\text {Volume in L}}=1.2M\times 0.01L=0.012mol

HCl+NaOH\rightarrow NaCl+H_2O

For neutralization:

1 mole of H^+ ion will react with 1 mole of OH^- ion

0.01 mol of H^+  ion will react with =\frac{1}{1}\times 0.01mole of OH^- ion

Thus (0.012-0.01)= 0.002 moles of OH^- are left in 20 ml or 0.02 L of solution.

[OH^-]=\frac{0.002}{0.02L}=0.1M

pOH=-log[OH^-]

pOH=-log[0.1]=1

pH+pOH=14

pH=14-1=13

Thus the pH of resulting solution will be 13

7 0
3 years ago
How many hybrid orbitals are found in CCl4? o one o two o three o four​
BlackZzzverrR [31]

Answer:

I Believe it is 4 orbitals s,p,p,p or aka sp^3

Explanation:

4 0
2 years ago
Read 2 more answers
Can you help me with this with a solution
LuckyWell [14K]

The complete table is inserted.

A table is given,

Formulas used:

pH=  -log(H⁺)

pOH=  -log(OH⁻)

pH+ pOH=14

Calculations:

For A: (H⁺)=2×10⁻⁸M

Using the pH formula:

pH=  -log(H⁺)=-log(2×10⁻⁸)=7.69

pOH=14 - 7.69=6.3

Calculating OH concentration,

pOH=  -log(OH⁻)

6.3= -log(OH⁻)

(OH⁻)=5.011×10⁻⁷M

Hence, the nature of A is basic.

Similarily,

For B,

(OH⁻)=1×10⁻⁷

Using the pH formula:

pOH=  -log(OH⁻)= -log(1×10⁻⁷)=7

pH=14-7=7

Calculating H concentration,

pH=  -log(H⁺)

7= -log(H⁺)

(H⁺)=1×10⁻⁷M

Hence, the nature of B is neutral.

Similarily,

For C,

pH=12.3

Using the pH formula:

pOH=14-12.3=1.7

Calculating H concentration,

pH=  -log(H⁺)

12.3= -log(H⁺)

(H⁺)=5.011×10⁻¹³M

Calculating OH concentration,

pOH=  -log(OH⁻)

1.7= -log(OH⁻)

(OH⁻)=1.99×10⁻²M

Hence, the nature of C is Basic.

Similarily,

For D,

pOH=6.8

Using the pH formula:

pH=14-6.8=7.2

Calculating H concentration,

pH=  -log(H⁺)

7.2= -log(H⁺)

(H⁺)=6.309×10⁻⁸M

Calculating OH concentration,

pOH=  -log(OH⁻)

6.8= -log(OH⁻)

(OH⁻)=1.58×10⁻⁷M

Hence, the nature of D is basic.

Learn more about the acid and bases here:

brainly.com/question/16189013

#SPJ10

3 0
2 years ago
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