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katen-ka-za [31]
3 years ago
12

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Chemistry
1 answer:
GenaCL600 [577]3 years ago
4 0

Answer:

Huh?

Explanation:

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An old penny rusting, remember you can’t undo a chemical change
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3 years ago
Type the correct answer in the box. Express your answer to three significant figures.
Shkiper50 [21]

Answer: Theoretical mass of sodium sulphate ( Na_2SO_4 ) is 514.118 grams.

Explanation: For a given reaction,

2NaOH(aq.)+H_2SO_4(aq.)\rightarrow Na_2SO_4(s)+2H_2O(l)

As NaOH is used in excess, therefore it is an excess reagent and H_2SO_4 is a limiting reagent as the quantity of the product will depend on it.

We are given 355 grams of H_2SO_4.

Molar mass of H_2SO_4 = 98.079 g/mol

Molar mass of Na_2SO_4 = 142.04 g/mol

1 mole of H_2SO_4 is producing 1 mole of Na_2SO_4, so

98.079 g/mol of  H_2SO_4 will produce 142.04 g/mol of Na_2SO_4

355 grams of H_2SO_4 will produce = 142.04g/mol \times \frac{355g}{98.079g/mol} of Na_2SO_4

Mass of Na_2SO_4 = 514.118 grams


3 0
4 years ago
Read 2 more answers
During a rainy spring, where are you most likely to find surface runoff
Dahasolnce [82]

Answer:On a sloped parking lot

Explanation:

7 0
3 years ago
Fill in the missing spaces in the chart. Will give brainliest.
lana [24]

Explanation:

9/4 Be +2 (the 9 and 4 are stacked next to Be). Atomic #: 4

Mass #: 9, # protons: 4, # neutrons: 5, #electrons: 2.

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Mass #: 31, # protons: 15, # neutrons: 16, # electrons: 15.

3 0
3 years ago
What is the concentration of a potassium iodate solution after you complete the following porcedure? Pipette 10 mL of a 0.31 M p
zubka84 [21]

<u>Given:</u>

Initial concentration of potassium iodate (KIO3) M1 = 0.31 M

Initial volume of KIO3 (stock solution) V1 = 10 ml

Final volume of KIO3 V2 = 100 ml

<u>To determine:</u>

The final concentration of KIO3 i.e. M2

<u>Explanation:</u>

Use the relation-

M1V1 = M2V2

M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M

Ans: The concentration of KIO3 after dilution is 0.031 M

4 0
3 years ago
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