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2 years ago

2.What two factors must be held constant for density to be a constant ratio?

Chemistry

2 answers:

irina1246 [14]2 years ago

6 0

Answer:

Temperature and Pressure

Explanation:

Temperature and pressure cause change in volume.

So any change in volume will alter the ratio of density as given by equation of density.

Density = mass/ volume

Change in volume will alter the ratio.

Kindly mark it branliest if the answer is little bit satisfying.

Drupady [299]2 years ago

5 0

The two (2) factors which must be held constant for density to be a constant ratio are:

I. The temperature of a sample or physical substance.

II. The pressure around the sample or physical substance.

Density can be defined as mass all over the volume of of a sample or physical substance.

Basically, density is mass per unit volume of of a sample or physical substance.

Mathematically, the density of a of a sample or physical substance is given by the formula;

$Density = \frac{Mass}{Volume}$

The density of a sample or physical substance is directly proportional to pressure and inversely proportional to temperature.

This ultimately implies that, an increase in pressure with temperature being constant would cause an increase in density and vice-versa.

Conversely, a decrease in temperature with pressure being constant would cause an increase in density and vice-versa.

In conclusion, a change in volume of a substance is caused by both temperature and pressure.

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It is known that neutral elements with same number of valence electrons will have same chemical properties. Therefore, elements of same group show similar chemical properties.

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3 years ago

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3 years ago

½O2(g) + H2(g) ⇌ H2O(g)

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Answer:

-241.826 kJ·mol⁻¹; -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous

Explanation:

½O₂(g) + H₂(g) ⟶ H₂O(g)

ΔHf°/kJ·mol⁻¹: 0 0 -241.826

S°/J·K⁻¹mol⁻¹: 205.0 130.6 188.7

1. ΔᵣH

ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹

3. ΔS(univ)

$\begin{array}{rcl}\Delta S_{\text{univ}} &=& \Delta S_{\text{sys}} +\Delta S_{\text{surr}}\\\\ &=& \Delta S_{\text{sys}} -\dfrac{\Delta H_{\text{sys}}}{T}\\\\& = & -146.9 - \dfrac{-241826}{298}\\\\& = & -146.9 + 811.5\\& = & \mathbf{664.6 \,\, J\cdot K^{-1}mol^{-1}}\\\end{array}$

4. Spontaneity

$\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -241.826 - 298 \times (-0.1469)\\& = & -241.826 + 43.776\\& = & \textbf{-198.050 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}$

ΔG is negative, so the reaction is spontaneous.

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3 years ago

An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a

Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):

$V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol$

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

$n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol$

$V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21$

$P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm$

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.

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2 years ago