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Liono4ka [1.6K]
1 year ago
11

What is the net ionic equation for this reaction?

Chemistry
1 answer:
Elenna [48]1 year ago
8 0

The ionic equation represents the element in cation and anionic form. The equation is given as OH⁻ + 2H⁺ → 2H₂O. Thus, option A is correct.

<h3>What is the ionic equation?</h3>

The ionic reaction equation is the representation of the chemicals dissociated as the ions in an aqueous solution. The ionic reaction is given as the cations and anions dissociated from the respective molecule or compound.

The ionic equation between potassium hydroxide and sulfuric acid is shown as,

2K⁺ + OH⁻ + 2H⁺ + SO₄²⁻ → 2H₂O + 2K⁺ + SO₄²⁻

It can also be written as,

Net ionic equation: OH⁻ + 2H⁺ → 2H₂O

Here, others are eliminated as they are spectator ions.

Therefore, option A. OH⁻ + 2H⁺ → 2H₂O is the net ionic equation.

Learn more about ionic equation here:

brainly.com/question/27980783

#SPJ1

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Bonjour,

increasing temperature.
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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

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In 1787,_________discovered a relationship between charge, their separation, and the force between the two charges. A coulomb B
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The answer is A. Columb

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How many grams of NaHCO3 are needed to prepare 250 mL of 0.50 M<br> NaHCO3?
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Answer:

10.5g

Explanation:

First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:

Volume = 250mL = 250/1000 = 0.25L

Molarity = 0.5M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.5 x 0.25

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Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:

Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol

Number of mole of NaHCO3 = 0.125 mole

Mass of NaHCO3 =?

Mass = number of mole x molar Mass

Mass of NaHCO3 = 0.125 x 84

Mass of NaHCO3 = 10.5g

Therefore, 10.5g of NaHCO3 is needed.

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