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Brums [2.3K]
3 years ago
10

A rigid vessel filled to 0.8 of its volume with liquid nitrogen at its normal boiling point is allowed to warm to 25°C. What pre

ssure is developed? The molar volume of liquid nitrogen at its normal boiling point is 34.7 cm3·mol–1.
Chemistry
1 answer:
rodikova [14]3 years ago
5 0

Answer:

704.25 atm

Explanation:

From the ideal gas equation

PV = nRT

P = nRT/V

Molar volume (V/n) = 34.7 cm^3/mol, n/V = 1/34.7 = 0.0288 mol/cm^3

R = 82.057 cm^3.atm/mol.K

T = 25°C = 25+273 = 298 K

P = 0.0288 × 82.057 × 298 = 704.25 atm

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Luba_88 [7]
An atom hopefully this helps
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3 years ago
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What is the density (g / mL) of a substance that has a mass of 8.11 g and, when placed into a graduated cylinder, causes the wat
MAXImum [283]

Answer:

<h3>The answer is 0.42 g/mL</h3>

Explanation:

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

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

From the question

mass = 8.11 g

volume = final volume of water - initial volume of water

volume = 44.72 - 25.26 = 19.46 mL

We have

density =  \frac{8.11}{19.46}  \\  = 0.416752312...

We have the final answer as

<h3>0.42 g/mL</h3>

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8 0
3 years ago
A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

7 0
3 years ago
Level 1: 2Sr + 02 &gt; 2Sro
tekilochka [14]

Answer:

Sr would be the limiting reactant

5 moles

Explanation:

Since the equation is a balanced equation, the coefficient shows how each substance relates to the other in terms of the number of moles.

Reactants would be those on the left hand side of the arrow, while the products would be found on te right and side of the arrow. In this question, the reactants would be Sr and O₂.

Limiting reactant is the reactant that is insufficient; meaning to say that there is not enough of that substance and thus the reaction cannot continue. The other reactant(s) that is not limiting is called the excess reactants.

From the balanced equation, 2 moles of Sr is needed to react with 1 mole of O₂. Thus, if we have 5 moles of each reactant, Sr would be the limiting reactant since for every 1 mole of O₂, there has to be 2 moles of Sr in order for the reaction to proceed. Thus, if we have 5 moles of O₂, we would need 10 moles of Sr.

When we work out the amount of products formed, we look at the number of moles of the limiting reactant. This is because the limiting reactant determines how much is being reacted, while the excess number of moles of the excess reactant will remain unreacted.

For every 2 moles of Sr reacted, 2 moles of SrO would be produced. This means that the mole ratio of Sr to SrO is 1:1. Thus, since 5 moles of Sr has been reacted, 5 moles of the product (SrO) would be produced.

8 0
3 years ago
25 POINTS:)
Digiron [165]
I suppose that the answer is A
6 0
3 years ago
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