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3 years ago

Mg(OH)2+2 HCI = MgCI2+ H2O

Chemistry

1 answer:

In-s [12.5K]3 years ago

7 0

Answer:

Mg(OH)2 + 2HCI ⇒ MgCI2 + 2H2O

Explanation:

In order to balance a chemical equation you need to make sure both sides of the equations are equal.

Mg(OH)2 + 2HCI = MgCI2 + H2O

Mg = 1

Oh = 2

HCI = 2

Products:

Mg = 1

CI = 2

H = 2

O = 1

2H20 = 1 × 2 = 2

2 × 2 = 4

2HCI

1 × 2 = 2

Mg(OH)2 + 2HCI ⇒ MgCI2 + 2H2O

Hope this helps.

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Answer:

Final concentration of NaOH = 0.75 M

Explanation:

For $NaOH$ :-

Given mass = 90.0 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{90.0\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 2.2502\ mol$

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.

The expression for the molarity, according to its definition is shown below as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Where, Volume must be in Liter.

It is denoted by M.

Given, Volume = 3.00 L

So,

$Molarity=\frac{2.2502\ mol}{3.00\ L}=0.75\ M$

<u>Final concentration of NaOH = 0.75 M</u>

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3 years ago

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Answer:

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A 2.500g sample of compound containing only Carbon and Hydrogen is found containing 2.002g of Carbon. Determine the empirical fo

Dima020 [189]

The empirical formula : CH₃

<h3>Further explanation</h3>

Given

2.5 g sample

2.002 g Carbon

Required

The empirical formula

Solution

Mass of Hydrogen :

= 2.5 - 2.002

= 0.498

Mol ratio C : H :

C : 2.002/12 = 0.167

H : 0.498/1 = 0.498

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2 years ago

A bucket gets filled faster at ground floor than at 2nd floor

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Answer:

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Explanation:

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3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as a buffer made from 475 mL of 0.

Hitman42 [59]

Explanation:

The given data is as follows.

[HCOOH] = 0.2 M, [NaOH] = 2.0 M,

V = 500 ml, [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

No. of moles of benzoic acid = Molarity × Volume

= $2 \times 0.475$

= 0.095 mol

And, moles of NaOH present in the solution will be as follows.

No. of moles of NaOH = Molarity × Volume

= $2 \times 0.025$

= 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

$C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O$

Initial: 0.095 0.05 0 0

Equlbm: (0.095 - 0.05) 0 0.05

pH = $pK_{a} + log \frac{Base}{Acid}$

= $4.2 + log \frac{0.05}{0.045}$

= 4.245

For,

$HCOOH + NaOH \rightarrow HCOONa + H_{2}O$

Initial: 0.2x 2(0.5 - x) 0

Equlbm: 0.2x - 2(0.5 - x) 0 2(0.5 - x)

As,

pH = $pK_{a} + log \frac{Base}{Acid}$

4.245 = 3.75 + $log \frac{Base}{Acid}$

$log \frac{Base}{Acid}$ = 0.5

$\frac{Base}{Acid}$ = 3.162

Now,

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}$ = 3.162

x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

= 0.036 L

= 36 ml (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

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3 years ago