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Leona [35]
2 years ago
13

How many grams of Ca(OH)2 are needed to produce 500 mL of 2.99 M Ca(OH)2 solution? 1L=1000mL

Chemistry
1 answer:
Alex_Xolod [135]2 years ago
3 0

Answer:

74 g/mol

Explanation:

Using a periodic table, we can determine the molar mass by adding together 1 Ca, 2 O, and 2 H. This turns out to be approximately (40+32+2) = 74 g/mol

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20 ML of a gas at 200 K is heated until the new volume is 55ML what is the final temperature of the gas
MrRa [10]

Answer:

T2 = 550K

Explanation:

From Charles law;

V1/T1 = V2/T2

Where;

V1 is initial volume

V2 is final volume

T1 is initial temperature

T2 is final temperature

We are given;

V1 = 20 mL

V2 = 55 mL

T1 = 200 K

Thus from V1/T1 = V2/T2, making T2 the subject;

T2 = (V2 × T1)/V1

T2 = (55 × 200)/20

T2 = 550K

8 0
2 years ago
A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionizati
aleksandr82 [10.1K]

Answer:

0.42%

Explanation:

<em>∵ pH = - log[H⁺].</em>

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

<em>∵ [H⁺] = √Ka.C</em>

∴ [H⁺]² = Ka.C

∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.

<em>∵ Ka = α²C.</em>

Where, α is the degree of dissociation.

<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>

<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>

4 0
3 years ago
Read 2 more answers
What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

3 0
3 years ago
If D+2 would react with E-", what do you predict to be the formula?
GuDViN [60]

Answer:DE2

Explanation:

6 0
3 years ago
Select two correct answers
Anna [14]

Answer:

D and E

Explanation:

The other answers don't support steel, they support iron or are supporting both, and the question is why steel alloys are more often used

C supports both

B supports how iron's strength

A supports iron being easily shaped and bent

hope this helps

6 0
2 years ago
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