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rosijanka [135]
3 years ago
10

4. Which metals will be able to reduce copper ions in solution?

Chemistry
1 answer:
topjm [15]3 years ago
3 0
4. <span>The metals that would be able to reduce copper ions in solution would be hydrogen(H), lead(Pb), tin(Sb), nickel(Ni), iron(Fe), zinc(Zn), aluminum(Al), Magnesium(Mg), sodium(Na), calcium(Ca), potassium(K), and lithium(Li).

5. </span>If you had a house with both copper- and zinc-galvanized iron water pipes,zinc would be desirable because <span>A metal that is easily oxidized would rust more readily.</span><span>

</span>
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The total bonding energy for the products of a reaction is 2535 kJ/mol and the bonding energy of the reactants is 1375 kJ/mol. C
Dmitry [639]

Answer:

-1160kj/mol

Explanation:

the reaction is exothermic because heat is released to the environment

7 0
3 years ago
Calculate the χacetone and χcyclohexane in the vapor above the solution. p°acetone = 229.5 torr and p°cyclohexane = 97.6 torr.
Ulleksa [173]
First calculate the mole fraction of each substance:
Acetone: 2.88 mol ÷ (2.88 mol + 1.45 mol) = 0.665
Cyclohexane: 1.45 ÷ (2.88 mol + 1.45 mol) = 0.335
Raoult's Law: P(total) = P(acetone) · χ(acetone)  + P(cyclohexane) · χ(cyclohexane).
P(total) = 229.5 torr · 0.665 + 97.6 torr · 0.335
P(total) = 185.3 torr
χ for acetone: 229.5 torr · 0.665 ÷ 185.3 torr = 0.823 
χ for cyclohexane:  97.6 torr · 0.335 ÷ 185.3 torr = 0.177
7 0
3 years ago
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To what extent do covalent compounds conduct electricity?
sertanlavr [38]

The answer is never because there's no electricity

3 0
3 years ago
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What is the chemical test to differentiate ethanol from ethanoic acid
Tomtit [17]

Answer:

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8 0
2 years ago
A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipita
mestny [16]

Answer:

The answer is:

(a) NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 0.030

=0.0015 \ mol

(a)

The reaction is:

⇒ NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)

(b)

1 mole of NaCl react with,

= 1 mol of AgNO_3

0.0015 mol AgNO_3 needs,

= 0.00150 \ mol \ NaCl

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

= 0.00147\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

4 0
3 years ago
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