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3 years ago

10

Which phrases describe space observatories?

1 answer:

amid [387]3 years ago

7 0

Answer:1 and 3

Explanation: got it right

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Which of the following shows correctly an ion pair and ionic compound the two ions from

zubka84 [21]

Answer:

what are the choices?

Explanation:

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3 years ago

4) list 3 protest rights you have that are protected under the law. How can you protect these rights? George Floyd’s ?

Pavel [41]

Peaceful Protest, Signing Pettitons, contacting mayors/governors and writing emails/sending letters

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3 years ago

Liquid does not act that a solid. how is it different from a solid and gas?

Wittaler [7]

Unlike solid matter, where particles are tightly packed and slightly vibrating, or gas, where particles go around everywhere and are extremely loose, a liquid has particles that are loosely packed but are still in slight contact with each other. Hope that's good enough

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

7 0

3 years ago

How many grams of iron(II) chloride are needed to produce 44.3 g iron(II) phosphate in the presence of excess sodium phosphate?

zalisa [80]

Answer:

47.2 g

Explanation:

Let's consider the following double displacement reaction.

3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl

The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

44.3 g × (1 mol / 357.48 g) = 0.124 mol

The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

3 × 0.124 mol = 0.372 mol

The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

0.372 mol × (126.75 g/mol) = 47.2 g

5 0

3 years ago