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3 years ago

Please help I soon as possible it’s really important

Chemistry

1 answer:

Vlada [557]3 years ago

8 0

Answer:

II. Rearrange the following jumbled words/phrases to make meaningful sentences: 5x1=5

a. certainly/the laughter/is/medicine/best

b. laughter/fact/is/in/a/panacea/for/diseases/all

c. lowers/pressure/by/circulation/it/blood/increasing/blood

d. a sense/and promotes/of/stress/well-being/laughter/decreases.

e. healthy food/ needed/is/better/also/lead/to/a/life

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Firewood burns and ashes remain<br><br> is it a chemical or physical change?

erastova [34]

It is a physical change because you can not put it back like it was

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3 years ago

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N2(g) + 3H2(g) => 2NH3(g) Write the reaction again as a balanced reaction

Evgen [1.6K]

Answer:Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). If hydrogen gas is added to this system at equilibrium, will the reaction shift towards reactants

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2 years ago

I need the answer to 452 of arogon

yaroslaw [1]

I'm confused... is there more info?

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3 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

PLEASEEEEEEEEEEEE HELPPPPPPPP I BEGGGGG FOR HELPPPPP

Elza [17]

Answer: There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{63.00g}{18g/mol}=3.5moles$

1 mole of $H_2O$ contains = $6.023\times 10^{23}$ molecules

Thus 3.5 moles of $H_2O$ contains = $\frac{6.023\times 10^{23}}{1}\times 3.5=21.08\times 10^{23}$ molecules.

There are $21.08\times 10^{23}$ molecules in 63.00 g of $H_2O$

3 0

3 years ago