It is so because warm air is lighter than cold air. Cold air means the molecules are not moving as much, so they will feel like sticking together and will not jump around as molecules in warm air will(higher temp. means higher KE). Think of a solid, colder solids are much heavier than warmer ones(think of ice-starts melting).

3 0

4 years ago

What’s Silicon’s melting point?

torisob [31]

Answer:

2,577°F

Also,

Boron: 3,769°F (2,076°C)

Neon: -415.5°F (-248.6°C)

Beryllium: 2,349°F (1,287°C)

8 0

3 years ago

Read 2 more answers

he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o

frosja888 [35]

<u>Answer:</u> The number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

<u>Explanation:</u>

We are given:

$K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L$

To calculate the concentration of a substance, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ ......(1)

Concentration of ammonia:

$[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L$

Concentration of ammonium iodide:

$[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L$

For the given chemical reaction:

$NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI][NH_3]}{[NH_4I]}$

Putting values in above equation, we get:

$7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}$

$[HI]=2.53\times 10^{-4}$

Calculating the moles of hydrogen iodide by using equation 1, we get:

$2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}$

Hence, the number of moles of HI in the solution is $1.24\times 10{-3}$ moles.

3 0

3 years ago

Which of the following happens when a reaction reaches dynamic equilibrium in a closed system?

laiz [17]

Answer: B

Explanation:

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and thus the concentrations of the reactants and products must be constant.

8 0

2 years ago