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2 years ago

Calculate the percentage of zinc I zncl2

Chemistry

1 answer:

Murrr4er [49]2 years ago

4 0

Answer:

47.8 %

Explanation:

Ar of Zn = 65

Mr of ZnCl2 = 65 + (2 * 35.5)

= 65+ 71

= 136

Percentage Composition

= ( Ar of element × No of atom in formula ) / Mr of compound * 100 %

% composition of Zn = ( 65*1 )/ 136 × 100 %

= 47.7941...

= 47.8 % ( 3 s.f )

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How many electrons are in a ion that has 16 protons and 59 neutrons if it has negative 2 charge?

GalinKa [24]

Hi there I believe it’s 18 please let me know if I’m wrong :)

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1 year ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

If 50.0 g of water saturated with potassium chloride at 80.0°C is slowly evaporator to dryness, how many grams of the dry salt w

balu736 [363]

Answer:

25.0 g.

Explanation:

From the solubility curve that is shown in the attached image, the solubility of KCl per 100.0 g of water is about 50.0 g.

So, a saturated solution of KCl in 50.0 g of water will contain about 25.0 g.

Thus, the grams of the dry salt that will be recovered is 25.0 g.

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3 years ago

Which of the following statements is true of combustion reactions?

enot [183]

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ΔG < 0, so the free energy of the system decreases with the reaction. Remember that when there is a negative ΔG the reaction goes from higher free energy to lower free energy, like in this case.

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kodGreya [7K]

567000 would be in grams, hope this helps :)

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3 years ago

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