1.Salts consist of atomic particles that have an (blank) charge.

Iteams in house made of thermosetting plastics?

Thermoplastics and thermosetting polymers Examples include: polyethylene (PS) and polyvinyl choline (PVC). Common thermoplastics range from 20,000 to 50,000 amu, while thermosets are assumed to have infinite molecular weight.

3 0

3 years ago

Given that hclo4 is a strong acid, how would you classify the basicity of clo−4?

Mandarinka [93]

As a conjugate base of a strong acid,ClO4-would be classified as having a negligible basicity. The basicity of a chemical species is normally expressed by the acidity of the conjugate acid. The basicity of an acid is the number of hydrogen ions, which can be produced by one molecule of the acid.

8 0

3 years ago

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.

4 0

3 years ago

What is the reduction half-reaction for the reaction of zinc with hydrochloric acid?

german

2HCl (aq) + Zn²⁺ ⇒ ZnCl₂ + 2H⁺ Total Reaction

Leo goes Ger (Lose electron oxidize, Gain electron reduction)

Zn²⁺ + 2e⁻ ⇒ Zn Oxidation half-reaction

2H ⇒ 2H⁺ + 2e⁻ Reduction half-reaction

6 0

3 years ago

An example of a potassium compound containing both ionic and covalent bonds is?

gayaneshka [121]

It's a trick question because all of them contain both iconic and covalent bonds.

Hope it helps!! :) have fun

4 0

3 years ago