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3 years ago

The molar mass of hydrogen (H2) is 2.02 g/mol. A sample contains 4.00 mol of H2. What is the mass, in grams, of this sample

1 answer:

4 0

The mass in grams is 8.08 g of (H2). To figure this out, you would take 2.02 and multiply it by 4.00 to get the answer of 8.08.

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Please write the process (how soap is made?.......................

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Soap is created by mixing fats and oils with a base as opposed to detergent which is created by combining chemical compounds in a mixer.

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2 years ago

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Write an equation for the reaction that takes place during the laboratory preparation of dry hydrogen

olya-2409 [2.1K]

Answer:

Zn+2HCl→ZnCl2+H2

Explanation:

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc.

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3 years ago

How many moles of each element (C, H, and O) are present in a 100.0 g sample of<br> ethanol?

Anuta_ua [19.1K]

Answer:

2.173 moles of ethanol is presented in a 100.0g sample of ethanol .

Explanation:

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3 years ago

At equilibrium at 2500K, [HCl]=0.0625M and [H2]=[Cl2]=0.00450M for the reaction H2+Cl2 ⇌ HCl.

ASHA 777 [7]

Answer:

a. $H_2+Cl_2 \rightleftharpoons 2HCl$

b. K = 192.9

c. Products are favored.

Explanation:

Hello!

a. In this case, according to the unbalanced chemical reaction we need to balance HCl as shown below:

$H_2+Cl_2 \rightleftharpoons 2HCl$

In order to reach 2 hydrogen and chlorine atoms at both sides.

b. Here, given the concentrations at equilibrium and the following equilibrium expression, we have:

$K=\frac{[HCl]^2}{[H_2][Cl_2]}$

Therefore, we plug in the data to obtain:

$K=\frac{(0.0625)^2}{(0.00450)(0.00450)}\\\\K=192.9$

c. Finally, we infer that since K>>1 the forward reaction towards products is favored.

Best regards!

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2 years ago

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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2 years ago