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Which of these compounds would you expect

Veseljchak [2.6K]

ClBr, two nonmetals

Hope this helps you

6 0

2 years ago

While exploring a coal mine, scientists found plant fossils in the ceiling of the mine which had been preserved by an earthquake

ale4655 [162]

Answer:

11552.45 years

Explanation:

Given that:

Half life = 5730 years

$t_{1/2}=\frac{\ln2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{5730}\ years^{-1}$

The rate constant, k = 0.00012 years⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = 0.00012 years⁻¹

Initial concentration $[A_0]$ = 160.0 counts/min

Final concentration $[A_t]$ = 40.0 counts/min

Time = ?

Applying in the above equation, we get that:-

$40.0=160.0e^{-0.00012\times t}$

$e^{-0.00012t}=\frac{1}{4}$

$-0.00012t=\ln \left(\frac{1}{4}\right)$

$t=11552.45\ years$

7 0

2 years ago

Beginning with commercial grade hydrochloric acid, 1.00 • 10^2 mL of a 12.4 M HCl is added to water to bring the total volume if

MatroZZZ [7]

Concentration=mass/volume.

This should help.

3 0

2 years ago

In ADEF, the measure of F=90°, the measure of ZD=10, and EF = 94 feet. Find the

Flauer [41]

Answer:

answer. the measure. is f=90 the equbaliint is 180

Explanation:

yan ang sagot ko

3 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