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2 years ago

What is the total volume occupied by 132 g of CO2 at STP

Chemistry

2 answers:

ad-work [718]2 years ago

7 0

132g x (22.4L/44 grams) = 67.2L
The answer = 67.2 L
Hope it helps!
Puffin

Lilit [14]2 years ago

4 0

132 grams x (1 mol / 44 grams) = 3 moles
3 moles X (22.4 L/ 1 mol) = 67.2 L

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Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

$deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol$

b) To calculate the constant we have the following expression:

$lnK=-\frac{deltaG_{rxn} }{RT}$

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

$lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066$

c) The equilibrium pressure of O₂ over M is:

$K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm$

3 0

2 years ago

If the [H3O+] of a solution is 1x 10-8 mol/L the [OH-] is

Studentka2010 [4]

Answer:

(H30+)= 1x10^-6 M

Explanation:

Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:

1x 10-8 mol/L equals to1x10-8 M

(H3O+) x (OH-) = 1x10^-14

(H30+)x 1x 10^-8 =1x10^-14

(H30+)= 1x10^-14/1x 10^-8

(H30+)= 1x10^-6 M

6 0

3 years ago

can someone solve me some papers I don't have time and I need to submit today please I will make you brilliant

dem82 [27]

Answer:

submit the papers onto this platform and we will try to help out.

6 0

2 years ago

Volume of water displaced = 0.175 L Mass g Density g/cm3 Hardness: = Determine the unknown mineral. Unknown mineral: 7

erastovalidia [21]

Answer:

Mass: 981.0 g

Density: 5.61 g/cm^3

Hardness: = 2.5 - 3

Unknown material: Chalcocite

Explanation:

5 0

2 years ago

How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

$pH+pOH=14\\pOH=14-8.94=5.06$

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

Now we have to calculate the moles of $OH^-$ .

Formula used : $Moles=Concentration\times Volume$

$\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles$

For neutralization, equal number of moles of $H^+$ ions will neutralize same number of $OH^-$ ions.

$\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles$

As, $H_2SO_4\rightarrow 2H^++SO^{2-}_4$

From this reaction, we conclude that

2 moles of $H^+$ ion is given by the 1 mole of $H_2SO_4$

$3309.8\times 10^{-9}$ moles of $H^+$ ion is given by $\frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9}$ moles of $H_2SO_4$

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of $H_2SO_4$ × Molar mass of sulfuric acid

Mass of sulfuric acid = $(1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g$

Therefore, the mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

3 0

2 years ago