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2 years ago

Please help me which one

Chemistry

1 answer:

mel-nik [20]2 years ago

3 0

Non of the above because protons and neutrons don’t mix with each one there

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Simple machines make work easier. Most simple machines reduce the amount of effort force needed to move an object. This cart has

Kruka [31]

I'm pretty sure it's D

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3 years ago

If 200 g of water is heated from 20 C to 40 C, then what is the heat added in kilocalories (same as food calories)?

Maslowich

Answer: 4Kcal

Explanation:

H= mcø

M=200g

C= 1 cal/g/°c

Ø= 40-20=20°c

H= 200*1*20= 4000calories= 4Kcal

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3 years ago

Which substance can not be decomposed by a chemical change?<br> (1) AlCl3 (3) HI<br> (2) H2O (4) Cu

user100 [1]

Cooper is the substance that can't be decomposed by a chemical chAnge. So 4)

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3 years ago

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A certain food web is made up of turtles, pond weeds, water fleas, and tadpoles. Which organism is most likely the producer in t

Simora [160]

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2 years ago

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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 $9.5 \times 10^{-3} M$ .

7 0

3 years ago