A solution containing a nonvolatile solute will have

True or false when matter changes it’s state it is a physical change because it is the same substance the whole time.

Blababa [14]

Answer:

true

Explanation:

5 0

2 years ago

Name every block in mc

maks197457 [2]

Answer:

Andesite.
Bedrock.
Basalt. Smooth.
Blackstone.
Gilded.
Calcite.
Cobblestone.
Mossy.
Deepslate.
Diorite.

7 0

2 years ago

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How many moles of copper are needed to react with sulfur to produce 0.25 moles of copper sulfide

Setler [38]

Cu + S ---> CuS
by reaction 1 mol 1 mol
from the problem 0.25 mol 0.25 mol
0.25 mol Cu

6 0

3 years ago

Read 2 more answers

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

max2010maxim [7]

Answer:

The answer to your question is: Initial temperature of copper = 67.1°C

Explanation:

Data

mass Copper = 248 g

volume Water = 390 ml

T1 water = 22.6°C

T2 = 39.9°C

T1 copper = ?

Specific heat water = 1 cal/g°C

Specific heat copper = 0.092 cal/g°C

Formula copper water

Heat is negative for copper because it releases heat

- mCp(T2 - T1) = mCp(T2 - T1)

- (248)(39.9 - T1) = 390 (1)((39.9 - 22.6) Substitution

-9895.2 + 248T1 = 390(17.3) Simplification

-9895.2 + 248T1 = 6747

248 T1 = 6747 + 9895.2

248 T1 = 16642.2

T1 = 16642.2 / 248

T1 = 67.1 °C Result

6 0

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

7 0

2 years ago