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sleet_krkn [62]
3 years ago
5

A 5.0 L flask contains 0.60 g of 02 at a temperature of 22*C, what is the pressure inside the flask?

Chemistry
1 answer:
Kruka [31]3 years ago
4 0

Answer:

The pressure inside the flask is 0,09 atm

Explanation:

We calculate the number of moles of O2, we convert the Temperature into unit KELVIN and use the formula of the ideal gases to calculate the pressure:

Weight 1 mol of 02= 15,999g x2= 32g

32g---1mol 02

0,60g---x=(0,60gx1mol O2)/32g= 0,019mol O2

T(K)=273+22=295K

PV=nRT

Px5.0l= 0,019mol x 0,082 lxatm/Kxmolx295 K

P=(0,019mol x 0,082 lxatm/Kxmolx295 K)/5.0l

P=0,09atm

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If volumes are additive and 253 mL of 0.19 M potassium bromide is mixed with 441 mL of a potassium dichromate solution to give a
Alexxx [7]

Answer:

The concentration of the Potassium Dichromate solution is 0.611 M

Explanation:

First of all, we need to understand that in the final solution we'll have potassium ions coming from KBr and also K2Cr2O7, so we state the dissociation equations of both compounds:

KBr (aq) → K+ (aq) + Br- (aq)

K2Cr2O7 (aq) → 2K+ (aq) + Cr2O7 2- (aq)

According to these balanced equations when 1 mole of KBr dissociates, it generates 1 mole of potassium ions. Following the same thought, when 1 mole of K2Cr2O7 dissociates, we obtain 2 moles of potassium ions instead.

Having said that, we calculate the moles of potassium ions coming from the KBr solution:

0.19 M KBr: this means that we have 0.19 moles of KBr in 1000 mL solution. So:

1000 mL solution ----- 0.19 moles of KBr

253 mL solution ----- x = 0.04807 moles of KBr

As we said before, 1 mole of KBr will contribute with 1 mole of K+, so at the moment we have 0.04807 moles of K+.

Now, we are told that the final concentration of K+ is 0.846 M. This means we have 0.846 moles of K+ in 1000 mL solution. Considering that volumes are additive, we calculate the amount of K+ moles we have in the final volume solution (441 mL + 253 mL = 694 mL):

1000 mL solution ----- 0.846 moles K+

694 mL solution ----- x = 0.587124 moles K+

This is the final quantity of potassium ion moles we have present once we mixed the KBr and K2Cr2O7 solutions. Because we already know the amount of K+ moles that were added with the KBr solution (0.04807 moles), we can calculate the contribution corresponding to K2Cr2O7:

0.587124 final K+ moles - 0.04807 K+ moles from KBr = 0.539054 K+ moles from K2Cr2O7

If we go back and take a look a the chemical reactions, we can see that 1 mole of K2Cr2O7 dissociates into 2 moles of K+ ions, so:

2 K+ moles ----- 1 K2Cr2O7 mole

0.539054 K+ moles ---- x = 0.269527 K2Cr2O7 moles

Now this quantity of potassium dichromate moles came from the respective  solution, that is 441 mL, so we calculate the amount of them that would be present in 1000 mL to determine de molar concentration:

441 mL ----- 0.269527 K2Cr2O7 moles

1000 mL ----- x = 0.6112 K2Cr2O7 moles = 0.6112 M

6 0
3 years ago
assuming nitrogen behaves like an ideal gas, what volume would 14.0 g of nitrogen gas (N2) occupy at STP? the gas constant is 0.
dimaraw [331]

Answer:

V = 22.41 L

Explanation:

Given data:

Mass of nitrogen = 14.0 g

Volume of gas at STP = ?

Gas constant = 0.0821 atm.L/mol.K

Solution:

Number of moles of gas:

Number of moles = mass/molar mass

Number of moles= 14 g/ 14 g/mol

Number of moles = 1 mol

Volume of gas:

PV = nRT

1 atm × V = 1 mol × 0.0821 atm.L/mol.K  × 273 K

V = 22.41 atm.L / 1 atm

V = 22.41 L

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3 years ago
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ArbitrLikvidat [17]

Answer:

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e) H2SO4 + 2(NaOH) -----> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) ----------------> CaCO3(s) + 2NH4NO3(aq)

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3 years ago
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Answer:

b

Explanation:

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Veronika [31]

Answer:

Explanation:

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