Wat two common uses for Zinc???

Chemistry

1 answer:

trapecia [35]2 years ago

5 0

Answer:

Zinc oxide is used in the manufacture of many products such as paints, rubber, cosmetics, pharmaceuticals, plastics, inks, soaps, batteries, textiles and electrical equipment.

Explanation:

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What mass of sodium sulfate is needed to make 2.5 L of 2.0 M solution?

Serjik [45]

D) 710 g

Step by Step:
Multiply 2.5 L by 2.0 M to solve for moles
2 mol/L • 2.5 L= 5 mol

Find formula weight of sodium sulfate

Na2SO4-142.04 g/mol

Na- 2(22.99)=45.98
S-32.06
O-4(16)=64

Multiply miles by formula weight

5 mole • 142.04 g/mol=710.2 g

7 0

3 years ago

A bottle of wine contains 0.21 moles of ethanol,C2H5OH. The volume of the solution is 0.1 L. Calculate the molarity.

Yuliya22 [10]

Answer:

2.1mol/L

Explanation:

Number of moles = 0.21 moles

Volume = 0.1L

Molarity of a substance is the number of moles of solute dissolved in a volume of solvent (L)

Molarity = number of moles / volume of solvent

Molarity = 0.21 / 0.1

Molarity = 2.1mol/L

8 0

3 years ago

How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium

Sedbober [7]

Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

HCl: 2 moles
Na: 1 mole
NaCl: 2 moles
H₂: 1 mole

You know the following masses of each element:

H: 1 g/mole
Cl: 35.45 g/mole
Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
Na: 23 g/mole
NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

HCl: 2 moles* 36.45 g/mole= 72.9 g
Na: 1 mole* 23 g/mole= 23 g
NaCl: 2 moles* 58.45 g/mole= 116.9 g
H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

$mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}$