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4 years ago

The reaction below demonstrates which characteristic of a base?

Chemistry

2 answers:

Alchen [17]4 years ago

7 0

Answer:

Answer is D

Explanation:

svlad2 [7]4 years ago

5 0

Answer: option D. the ability of a base to react with a soluble metal salt.

Justification:

NaOH is a strong base, which means that in water it will dissociate according to this reaction:

NaOH(aq) → Na⁺ (aq) + OH⁻ (aq)

On the other hand, CuSO₄ is a soluble ionic salt which in water will dissociate into its ions according to this other reaction:

CuSO₄(aq) → Cu²⁺ + SO₄²⁻

Hence, in solution, the sodium ion (Na⁺) will react with the metal salt in a double replacement reaction, where the highly reactive sodium ion (Na⁺) will substitute the Cu²⁺ in the CuSO₄ to form the sodium sulfate salt, Na₂SO₄ (water soluble), and the copper(II) hydroxide, Cu(OH)₂ (insoluble).

That is what the given reaction represents:

CuSO₄ (aq) + 2NaOH(aq) → Cu(OH)₂(s) + Na₂SO₄(aq)

↑ ↑ ↑ ↑

soluble metal salt strong base insoluble base solube salt

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Atoms that have a full outer level combine with other atoms. Is this true or false

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3 years ago

A sample of metal has a mass of 16.5916.59 g, and a volume of 5.815.81 mL. What is the density of this metal

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Answer:

=> 2.8554 g/mL

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m) = 16.59 g

Volume (v) = 5.81 mL

From our question, we are to determine the density (rho) of the rock.

The formula:

$p = \frac{m}{v}$

Substitute the values into the formula:

$p = \frac{16.59 g}{5.81 mL} \\ = 2.8554 g/mL$

= 2.8554 g/mL

Therefore, the density (rho) of the rock is 2.8554 g/mL.

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2 years ago

Carbon dioxide is produced by the human body through

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Answer:

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3 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

3 years ago