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3 years ago

Which equation agrees with the ideal gas law? mc005-1.jpg mc005-2.jpg mc005-3.jpg mc005-4.jpg

2 answers:

7 0

The ideal gas equation is pV = nRT

From that you can derive several equations, depending on which variables are fixed.

1) When n and T are fixed:

pV = nRT = constant

pV = constant => p1 V1 = p2 V2 => p1 / V2 = p2 / V1 ---> Boyle's Law

2) When n and V are constant:

p / T = nR/V = constant

p / T = constant => p1 / T1 = p2 / T2 ----> Gay - Lussac's Law

3) when n and p are constant

V / T = nR/p = constant

V / T = constant => V1 / T1 = V2 / T2 ---> Charles' Law

4) When only n is constant

pV / T = nR = constant

pV / T = constant => p1 V1 / T1 = p2 V2 / T2 ----> Combined gas law.

There you have the four equations that agree with the ideal gas law.

Yuliya22 [10]3 years ago

3 0

Answer:

V1/T1=V2/T2

v1/n1=v2/n2

P1V1=P2V2

P1/T1=P2/T2

Explanation:

its right on edg.

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What will be the effect on the solubility of BaF2 if the following changes are made: increasing temperature; adding NaF; adding

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In BaF₂ the solubility will decreases on adding NaF

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2 years ago

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3 years ago

A solid inorganic nitrate contains 9.37% magnesium 10.93% nitrogen oxygen and 42.18% water calculate formula of a compound.

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This problem is providing us with the percent composition of a compound containing magnesium, nitrogen, oxygen and water. Thus, the empirical formula is required and found to be Mg(NO₃)₂·6H₂O.

<h3>Empirical formulas</h3>

In chemistry, empirical formulas are used in order to figure out the minimum whole-number representation of a molecular formula, which contains the actual number of atoms.

Thus, we start determining it by assuming the given percentages as masses and calculating the moles with the atomic or molar mass of the involved species:

$n_{Mg}=\frac{9.37}{24.3} =0.386\\\\n_{N}=\frac{10.3}{14.01}=0.735\\ \\n_O=\frac{37.52}{16.00}=2.32\\ \\n_{H_2O}=\frac{42.18}{18.02}=2.34$

Next, we divide by the 0.386 as the smallest number of moles in order to determine their subscripts in the chemical formula:

$Mg}=\frac{0.386}{0.386} =1\\\\n_{N}=\frac{0.735}{0.386}=2\\\\n_O=\frac{2.32}{0.386}=6\\\\n_{H_2O}=\frac{2.34}{0.386}=6$

Hence, the empirical formula turns out to be Mg(NO₃)₂·6H₂O, which is actually a hydrate (hexahydrate).

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