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3 years ago

7

Rank the metals from most reactive (1) to least reactive (3).

2 answers:

Citrus2011 [14]3 years ago

8 0

Answer:

1:Silver

2:Strontium

3:Technetium

Hope this helps!! Good Luck!!

alisha [4.7K]3 years ago

6 0

Answer:

Silver: 3

Strontium: 1

Technetium: 2

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Since density depends on the mass and volume of an object, we need both of these values combined in the correct way to solve for

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Answer:

7.86 g/cm³

Explanation:

11.0 kg = 11,000 g

The density in g/cm³ is ...

(11,000 g)/(1,400 cm³) = 7.86 g/cm³

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3 years ago

Write a net ionic equation that describes a slightly basic solution resulting from dissolving nac2h3o2 in water. Identify all of

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Answer: The net ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + OH^{-}(aq)$

Explanation:

The chemical equation for the given reaction is as follows.

$CH_{3}COONa(aq) + H_{2}O(l) \rightleftarrow CH_{3}COOH(aq) + NaOH(aq)$

We know that a strong acid or base will dissociate completely into a solvent whereas a weak acid or base dissociates partially into the solvent. Hence, the ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + Na^{+}(aq) H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + Na^{+}(aq) + OH^{-}(aq)$ Now, we will cancel the spectator ions from the above equation. Therefore, the net ionic equation will be as follows.

$CH_{3}COO^{-}(aq) + H_{2}O(l) \rightleftharpoons CH_{3}COOH(aq) + OH^{-}(aq)$

or, $C_{2}H_{3}O^{-}_{2}(aq) + H_{2}O(l) \rightleftharpoons HC_{2}H_{3}O_{2}(aq) + OH^{-}(aq)$

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3 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

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<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

<u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

<u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

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$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

<u>Initial:</u> 0.1372 0.04514

<u>Final:</u> 0.09206 - 0.04514

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$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

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Answer:

A) 0 charge

Explanation:

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