Ask question

Ask question

All categories

2 years ago

10

Name:

1 answer:

romanna [79]2 years ago

5 0

Answer:

C. Similar wind patterns were observed in Australia and South America

Explanation:

Hope it helps.

You might be interested in

What happens to electrons in the photoelectric effect?

11111nata11111 [884]

The photoelectric effect occurs when light shines on a metal. ... Light of any frequency will cause electrons to be emitted.

5 0

3 years ago

Read 2 more answers

4.2g of sodium bicarbonate is equivalent to how many moles of sodium bicarbonate

Mnenie [13.5K]

<span>The mass of one mole of sodium bicarbonate (aka NaHCO3) is equal to 1 * 22.99g/mol + 1 * 1.00g/mol + 1 * 12.01g/mol + 3 * 16.00g/mol = 83.91g/mol. From this, we can convert 4.2g of NaHCO3 to moles by dividing by 83.91g/mol, to get 0.050 moles of sodium bicarbonate.</span>

7 0

3 years ago

Read 2 more answers

What element has 2 electrons in its outermost energy level and is in the 2nd period on periodic table?

Olegator [25]

Be-beryllium have 2 electrons and it is in the 2 nd period

4 0

3 years ago

A drinking water plant adds 500 grams of fluoride to a water tank containing 500,000 liters of drinking water. what is the conce

nlexa [21]

Answer is: <span>concentration of fluoride in the water in parts-per-million is 1 ppm.
</span>Parts-per-million (10⁻⁶) is<span> present at one-millionth of a </span>gram per gram of sample solution, f<span>or example mg/kg.
</span>m(fluoride) = 500 g · 1000 mg/g = 500000 mg.
m(water) = d(water) · V(water).
m(water) = 1 kg/L · 500000 L.
m(water) = 500000 kg.
arts-per-million = 500000 mg ÷ 500000 kg = 1 mg/kg = 1 ppm.

5 0

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

7 0

3 years ago