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2 years ago

Please answer as quick as possible

Chemistry

2 answers:

OverLord2011 [107]2 years ago

4 0

Answer:

A - potassium........ M

pantera1 [17]2 years ago

4 0

Answer:

POTASSIUM

Explanation:

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Does a radio have electrical and mechanical energy

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Answer:

Electrical

Explanation:

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2 years ago

The graph below shows how solubility changes with temperature

torisob [31]

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2 years ago

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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.

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2 years ago

Which separation method would be most successful in separating the components of a homogeneous mixture?

stiks02 [169]

I'm pretty sure the answer is B) Evaporation.

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3 years ago

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When 18.5 g of HgO(s) is decomposed to form Hg(l) and O2(g), 7.75 kJ of heat is absorbed at standard-state conditions. What is t

taurus [48]

Answer:

The standard enthalpy of formation of HgO is -90.7 kJ/mol.

Explanation:

The reaction between Hg and oxygen is as follows.

$\text{Hg(l)}+\frac{1}{2}{O_{2}\rightarrow \text{HgO(s)}$

From the given,

Molar mass of HgO = 216.59 g/mol

Mass of HgO decomposed = 18.5 g

Amount of heat absorbed = 7.75 kJ

From the reaction,

The standard enthalpy of formation = $+7.75\times\frac{kJ}{18.5 g}\frac{216.59}{1mol} \,\,= +90.7 kJ/mol$

During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.

For the formation of 1 mol of HgO , 90.7 kJ of energy is release

Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol

5 0

3 years ago