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3 years ago

Complete and balance the following neutralization reaction Ca(OH)2 + H2SO4

Chemistry

1 answer:

Debora [2.8K]3 years ago

6 0

Answer: Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + 2H2O(l)

Explanation:

Since this is a neutralization reaction, the end product would be salt and water. In this equation Calcium will displace hydrogen from the acid because it is more reactive, resulting in the formation of CaSO4 (salt), while the displaced H2 molecule combines with OH molecules to form water.

The equation of the reaction is thus;

Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + H2O(l), in other to balance it, we add ''2'' to the water molecule in the right hand side of the equation.

Balance equation is

Ca(OH)2 (aq) + H2SO4 (aq) ----------> CaSO4(aq) + 2H2O(l)

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3 years ago

Read 2 more answers

if only 0.264 g of ca(oh)2 dissolves in enough water to give 0.178 l of aqueous solution at a given temperature, what is the ksp

Maurinko [17]

The ksp value for calcium hydroxide at this temperature is 5.20 × 10⁻⁶.

Ksp is an equilibrium constant of a solid substance dissolved in a liquid solution.

Given that, the volume of water is 0.178 l, 0.264 g of Ca(OH)² dissolves in enough water

The solution equilibrium is

Ca(OH)² = Ca + 2OH

The molar solubility is

0.186 / 74.00 / 0.230 = 0.0109 M

The ksp value will be

Ksp = (s) (2S)²

Putting the values in the formula

(0.0109) x (2 x 0.0109)² = 5.20 × 10⁻⁶.

Thus, the ksp value for calcium hydroxide is 5.20 × 10⁻⁶.

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1 year ago

Write a paragraph about the three forms of energy involved in a muscle contraction (chemical, mechanical, and thermal).

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3 years ago

A sample of sodium sulfite has a mass of 2.80 g.

Gnom [1K]

Answer:

For a: The number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b: The number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c: The mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

Explanation:

The chemical formula of sodium sulfite is $Na_2SO_3$ . It is formed by the combination of 2 sodium $(Na^+)$ ions and 1 sulfite $(SO_3^{2-})$ ions

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium sulfite = 2.80 g

Molar mass of sodium sulfite = 126 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium sulfite}=\frac{2.80g}{126g/mol}=0.022mol$

For a:

Moles of sodium ions in sodium sulfite = (2 × 0.022) moles

According to mole concept:

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

So, 0.022 moles of sodium sulfite will contain = $(2\times 0.022\times 6.022\times 10^{23})=2.65\times 10^{22}$ number of sodium ions

Hence, the number of sodium ions in given amount of sodium sulfite are $2.65\times 10^{22}$

For b:

Moles of sulfite ions in sodium sulfite = (1 × 0.022) moles

According to mole concept:

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

So, 0.022 moles of sodium sulfite will contain = $(1\times 0.022\times 6.022\times 10^{23})=1.325\times 10^{22}$ number of sulfite ions

Hence, the number of sulfite ions in given amount of sodium sulfite are $1.325\times 10^{22}$

For c:

Molar mass of sodium sulfite = 126 g/mol

According to mole concept:

$6.022\times 10^{23}$ number of formula units are present in 1 mole of a compound

Or, $6.022\times 10^{23}$ number of formula units of sodium sulfite have a mass of 126 grams

So, 1 formula unit of sodium sulfite will have a mass of = $\frac{126}{6.022\times 10^{23}}\times 1=2.09\times 10^{-22}g$

Hence, the mass of one formula unit of sodium ions is $2.09\times 10^{-22}$ grams

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