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2 years ago

Why is cold air cold?

Chemistry

1 answer:

Sholpan [36]2 years ago

4 0

Explanation:

Energy makes molecules in the air move faster and expand, decreasing density. The opposite is true for cold air. Molecules are closer together, and absorb less energy.

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A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?

Shalnov [3]

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

5 0

2 years ago

What elements are being reduced and oxidized in this chemical formula?

dangina [55]

Answer: The element Na (Sodium) is getting oxidized and Hydrogen is getting reduced.

Explanation:

Oxidation reactions are the reactions in which addition of oxygen takes place.

Reduction reactions are the reactions in which loss of oxygen takes place.

For a given reaction:

$2Na(s)+2H_2O(l)\rightarrow H_2(g)+2NaOH(aq.)$

Sodium is getting oxidized because there is an addition of reaction with that element.

Hydrogen is getting reduced because there is a removal of oxygen with that element.

3 0

2 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

6 0

2 years ago

Determine the number of moles present in 32.5 g aluminum chloride.

Artyom0805 [142]

The number of moles in 32.5g of aluminum chloride is approximately 0.250 moles.

3 0

2 years ago

3. When two atoms of 2H (deuterium) are fused to form one atom of 4He (helium), the total energy evolved is 3.83 × 10-12 joules.

balandron [24]

3.
∆E = ∆m x c ² ∆m = E / c ² ∆m = 3,83•10^-12 / 3•10^8 ² ∆m = 4,256•10^-29 kg

Taking this class as well

6 0

3 years ago

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