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sergiy2304 [10]
3 years ago
8

If 355 mL of 1.50 M aluminum nitrate is added to an excess of sodium sulfate, how many grams of aluminum sulfate will be produce

d?
Chemistry
1 answer:
Dimas [21]3 years ago
7 0
Answer is: 91.1 grams <span>of aluminum sulfate.
</span>Balanced chemical reaction: 2Al(NO₃)₃ + 3Na₂SO₄→ Al₂(SO₄)₃ + 6NaNO₃.
V(Al(NO₃)₃) = 355 mL ÷ 1000 mL/L = 0.355 L.
c(Al(NO₃)₃) = 1.5 mol/L.
n(Al(NO₃)₃) = V(Al(NO₃)₃) · c(Al(NO₃)₃).
n(Al(NO₃)₃) = 0,355 L · 1.5 mol/L.
n(Al(NO₃)₃) = 0.5325 mol.
From chemical reaction: n(Al(NO₃)₃) : n(Al₂(SO₄)₃) = 2 : 1.
n(Al₂(SO₄)₃) = 0.266 mol.
m(Al₂(SO₄)₃) = 0.266 mol · 342.15 g/mol.
m(Al₂(SO₄)₃) = 91.1 g.
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Bromine, a liquid at room temperature has a boiling point of 58 degrees celsius and a melting point at -7.2 degree celsius bromi
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Hello!


Bromine can be classified as a pure substance.


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Bromine is an element with atomic number 35 on group 17 of the Periodic Table. That's the first sign that shows us that it is a pure substance.


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8 0
3 years ago
Ava threw a hamster at her sister the other day she used 12N of force to accelerate the hamster at 8m/s2.what was the mass of th
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<u>We are given:</u>

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<u>Solving for the mass of the Poor Hamster:</u>

From newton's second equation of motion, we know that:

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8 0
2 years ago
How many grams Glycylglycine (molecular mass 132.119 grams/mole) are required to make 1 liter at 0.5 Molar concentration. If you
SOVA2 [1]

Explanation:

The given data is as follows.

        Concentration of solution = 0.5 M

        Volume of solution = 1 L

Molar mass of Glycylglycine = 132.119 g/mol

As molarity is the number of moles present in liter of solvent.

Mathematically,      Molarity = \frac{\text{No. of moles}}{\text{Volume}}

Hence, calculate the number of moles as follows.

            No. of moles = Molarity × Volume

                                  = 0.5 M \times 1 L

                                  = 0.5 mol

Therefore, mass of glycylglycine = mol × molar mass

                                                      = 0.5 mol \times 132.119 g/mol

                                                      = 66.06 g

Thus, we can conclude that 66.06 g glycylglycine is required.

8 0
3 years ago
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