In a solution of water and table salt, what is the table salt?

What products would you expect from the reaction of ammonia and sulfuric acid in aqueous solution

Lemur [1.5K]

Answer:

(NH4)2SO4

Explanation:

2NH3(aq) + H2SO4 (aq) ---> (NH4)2SO4(aq)

Ammonia reacts with sulfuric acid to give the fertilizer ammonium sulfate.

6 0

2 years ago

How many grams of water can be formed when 80 grams of sodium hydroxide (NaOH) reacts with an excess of sulfuric acid, H2SO4, in

podryga [215]

Answer:

36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction:

2NaOH + H2SO4 —> Na2SO4 + 2H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g.

From the balanced equation above, we can see evidently that:

80g of NaOH reacted to produce 36g of H2O.

5 0

4 years ago

How many atoms of phosphorus are in 7.00 mol of copper(II) phosphate?

Rama09 [41]

<u>Answer:</u> The number of phosphorus atoms in given amount of copper(II) phosphate is $8.431\times 10^{24}$

<u>Explanation:</u>

We are given:

Moles of copper(II) phosphate $Cu_3(PO_4)_2$ = 7.00 mol

1 mole of copper(II) phosphate contains 3 moles of copper, 2 moles of phosphorus and 8 moles of oxygen atoms

Moles of phosphorus in copper(II) phosphate = $(2\times 7.00mol$

According to the mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 7.00 moles of copper(II) phosphate will contain = $(2\times 7\times 6.022\times 10^{23}=8.431\times 10^{24}$ number of phosphorus atoms.

Hence, the number of phosphorus atoms in given amount of copper(II) phosphate is $8.431\times 10^{24}$

3 0

3 years ago

Chlorofluorocarbons such as CCl2F and CCI,F, have been linked to ozone depletion in Antarctica. In 1994, these gases were found

Ilya [14]

Answer:

a) molarity of CCl3F = 1.12 × 10^-11 mol/dm³

Molarity of CCl2F2 = 2.20 × 10^-11 mol/dm³

B) molarity of CCL3F = 7.96 × 10 ^-13 mol/dm³

Molarity of CCl2F2 = 1.55 × 10^-12 mol/dm³

Explanation:

Using the ideal gas equation:

PV = nRT

Further explanations are found in the attachment below.

4 0

4 years ago

A saturated solution of Al(OH)3 has a molar solubility of2.9x J0-9 Mat a certain temperature. What is the solubility product con

Viktor [21]

The solubility product constant, Ksp, of the aluminum hydroxide, Al(OH)₃ at the temperature is 1.91×10¯³⁹

<h3>What is solubility of product? </h3>

The solubility of product (Ksp) is defined as the concentration of products raised to their coefficient coefficients. This is illustrated below:

mA <=> nC + eD

Ksp = [C]^n × [D]^e

With the above information in mind, we can obtain the solubility of the product. This is illustrated below:

<h3>Dissociation equation </h3>

Al(OH)₃(aq) → Al³⁺(aq) + 3OH¯(aq)

From the balanced equation above,

1 mole of Al(OH)₃ contains 1 mole of Al³⁺ and 3 moles of OH¯

<h3>How to determine the concentration of Al³⁺ and OH¯</h3>

From the balanced equation above,

1 mole of Al(OH)₃ contains 1 mole of Al³⁺ and 3 moles of OH¯

Therefore,

2.9×10¯⁹ M Al(OH)₃ will also contain

1 × 2.9×10¯⁹ = 2.9×10¯⁹ M Al³⁺
3 × 2.9×10¯⁹ = 8.7×10¯⁹ M OH¯

<h3>How to determine the solubility of product </h3>

Concentration of Al³⁺ = 2.9×10¯⁹ M

Concentration of OH¯ = 8.7×10¯⁹ M

Solubility product (Ksp) =?

Al(OH)₃(aq) → Al³⁺(aq) + 3OH¯(aq)

Ksp = [Al³⁺] × [OH¯]³

Ksp = 2.9×10¯⁹ × (8.7×10¯⁹)³

Ksp = 1.91×10¯³⁹

Learn more about solubility of product:

brainly.com/question/4530083

#SPJ1

5 0

3 years ago