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Masja [62]
3 years ago
9

Acid rain has a higher than normal pH. a. True b. False

Chemistry
1 answer:
MAXImum [283]3 years ago
4 0
False.  Since acids are proton donors and pH=-log[H⁺], acids lower the pH of the solution that they are put into.  Therefore acid rain will have a lower pH than normal.
I hope this helps.
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WS Percent yield don’t understand how to do would appreciate the help
Grace [21]

Answer:

1. Theoretical yield of NaOH is 22.72 g

2. Percentage yield of NaOH = 22.14%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaHCO₃ —> NaOH + CO₂

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 1 mole (i.e 40 g) of NaOH and 1 mole (i.e 44.01 g) of CO₂.

Next, we shall determine the number of mole of NaHCO₃ that will decompose to produce 25 g of CO₂. This can be obtained as follow:

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 44.01 g of CO₂.

Therefore, Xmol of NaHCO₃ will decompose to 25 g of CO₂ i.e

Xmol of NaHCO₃ = 25 / 44.01

Xmol of NaHCO₃ = 0.568 mole

1. Determination of the theoretical yield of NaOH.

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 40 g of NaOH.

Therefore, 0.568 mole of NaHCO₃ will decompose to produce = 0.568 × 40 = 22.72 g of NaOH.

Thus, the theoretical yield of NaOH is 22.72 g

2. Determination of the percentage yield of NaOH.

Theoretical yield of NaOH = 22.72 g

Actual yield of NaOH = 5.03 g

Percentage yield of NaOH =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.03 / 22.72 × 100

Percentage yield of NaOH = 22.14%

4 0
2 years ago
If 10.0 grams of NaHCO3 is added to 10.0 g of HCl, determine the efficiency of baking soda as an antacid if 6.73 g of NaCl was p
Lapatulllka [165]

Answer:

percentage yield of NaCl = 96.64%

Explanation:

The reaction was between NaHCO3 and HCl .The chemical equation can be represented below:

NaHCO3 + HCl → NaCl + H2O + CO2 . The balance equation is

NaHCO3 + HCl → NaCl + H2O + CO2

The question ask us to calculate the percentage yield of NaCl.

The efficiency of NaHCO3 as an antacid , the limiting reactant is NaHCO3

as

1 mole of NaHCO3 produces 1 mole of NaCl

Therefore,

molar mass of NaHCO3 = 23 +1 + 12 + 48 = 84 g

molar mass of NaCl = 23 + 35.5 = 58.5 g

1 mole of NaHCO3 = 84 g

1 mole of NaCl  = 58.5 g

since 84 g of NaHCO3 produces 58.5 g of NaCl

10 g of NaHCO3 will produce ? grams of NaCl

cross multiply

Theoretical yield of NaCl = (10 × 58.5)/84

Theoretical yield of NaCl = 585/84

Theoretical yield of NaCl  = 6.9642857143 g

percentage yield of NaCl = actual yield/theoretical yield × 100

percentage yield of NaCl = 6.73/6.9642857143 × 100

percentage yield of NaCl = 673/6.9642857143

percentage yield of NaCl = 96.635897436%

percentage yield of NaCl = 96.64%

3 0
3 years ago
Read 2 more answers
Calculate the molarity of 0.850 mol of Na2S in 1.70 L of solution.
Basile [38]

Molarity is simply the ratio of the number of moles of a substance over the total volume of the solution. Assuming that the addition of 0.850 moles does not change the overall volume of 1.70 L, therefore molarity is simply:

 

Molarity = 0.850 moles / 1.70 L = 0.5 moles / L = 0.5 M

5 0
3 years ago
A gas mixture with a total pressure of 765 mmHg contains each of the following gases at the indicated partial pressures: 134 mmH
ladessa [460]

Answer: c

Explanation:

4 0
3 years ago
What is used to determine the number of each atom in an ionic formula
ladessa [460]

Answer:

The charge carried by each ion (oxidation state of each atom)

Explanation:

If we have an ionic compound and we want to write its formula, we must first know the magnitude of charge on each ion (shown as oxidation state of the atoms involved) because the magnitude of charge on each ion is eventually crisscrossed and gives the subscript (number of atoms) for each atom in the formula.

For instance, let us write the formula of calcium bromide. Ca has a charge of +2 while Br has a charge of -1. If we exchange the charges and ignore the signs such that the crisscrossed charges form subscripts we can now write; CaBr_{2}.

3 0
2 years ago
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