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Vaselesa [24]
3 years ago
9

True or false: nitrogen gas behaves more like an ideal gas as the temperature increases

Chemistry
1 answer:
navik [9.2K]3 years ago
5 0

Is true. Nitrogen gas behaves more like an ideal gas as the temperature increases. Under normal conditions such as normal pressure and temperature conditions , most real gases behave qualitatively as an ideal gas. Many gases such as air , nitrogen , oxygen ,hydrogen , noble gases , and some heavy gases such as carbon dioxide can be treated as ideal gases within a reasonable tolerance. Generally, the removal of ideal gas conditions tends to be lower at higher temperatures and lower density (that is at lower pressure ), since the work made by the intermolecular forces is less important compared to the kinetic energy<span> of the particles, and the size of the molecules is less important compared to the empty space between them. </span><span>The ideal gas model tends to fail at lower temperatures or at high pressures, when intermolecular forces and intermolecular size are important.</span>

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Does salt affect how long an ice cube melts?
Mashutka [201]

Answer:

Ice cubes with salt melt faster. When you add salt, it dissolves in the water of the ice cube. The freezing temperature of salt water is 32 degrees Fahrenheit lower than that of fresh water.

5 0
2 years ago
¿Cuál era su juego favorito cuando eran niños?​
blagie [28]

Answer:

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Explanation:

7 0
3 years ago
I want to know the steps.
Artyom0805 [142]

The answer for the following problem is described below.

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

Explanation:

Given:

enthalpy of combustion of glucose(ΔH_{f} of C_{6}H_{12} O_{6}) =-1275.0

enthalpy of combustion of oxygen(ΔH_{f} of O_{2}) = zero

enthalpy of combustion of carbon dioxide(ΔH_{f} of CO_{2}) = -393.5

enthalpy of combustion of water(ΔH_{f} of H_{2} O) = -285.8

To solve :

standard enthalpy of combustion

We know;

ΔH_{f}  = ∈ΔH_{f} (products) - ∈ΔH_{f} (reactants)

C_{6}H_{12} O_{6} (s) +6 O_{2}(g) → 6 CO_{2} (g)+ 6 H_{2} O(l)

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]

ΔH_{f} = [6 (-393.5) + 6(-285.8)] - [0 - 1275]

ΔH_{f} = 6 (-393.5) + 6(-285.8)  - 0 + 1275

ΔH_{f} = -2361 - 1714 - 0 + 1275

ΔH_{f} =-2800 kJ

<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>

7 0
2 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
3 years ago
How to solve for enthalpy for changes​
Natasha_Volkova [10]
Always use this method !!! always

7 0
3 years ago
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