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Bingel [31]
4 years ago
9

How much concentrated solution would you take to prepare 1.15 L of 0.105 M HNO3 by mixing with water?

Chemistry
1 answer:
nydimaria [60]4 years ago
7 0

Answer:

7.67578 ml

Explanation:

assuming the missing information

Conc. of nitric acid = 70.3% by mass

density =1.41 g/ml

Now, moles of HNO3 required = final volume× final conc.

= 1.15×0.105= 0.12075 mol

Mass of HNO3 required = moles×molar mass of HNO3

= 0.12075×63.01=7.6084575 g

mass of solution = 100/70.3 of 7.6084575

=10.82284 g

therefore, volume of the solution = mass/density

= 10.82285/1.41

= 7.67578 ml

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What is the answer for chemestry when it says 1.204 times 10^24
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Answer:

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8 0
3 years ago
How many minutes will it take for the quantity of n2o5 to drop to 1.9×10−2 mol ?
Novay_Z [31]

<span>Missing question: The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g) + O2(g) at 70°</span><span>C is 6.82×10−3 s−1. Suppose we start with 2.70×10−2 mol of N2O5(g) in a volume of 1.8 L .
</span>c₀(N₂O₅) = 0,027 mol ÷ 1,8 L.<span>
c</span>₀(N₂O₅) = 0,015 mol/L.<span>
c(N</span>₂O₅) = 0,019 mol/ 1,8 L = 0,01055 mol/L.<span>
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3 0
3 years ago
a solution is made by dissolving cadmium fluoride (CdF2 Ksp =6.44×10 −3 ) in 100.0 mL of water until excess solid is present. So
Lunna [17]

Answer:

Answers are in the explanation

Explanation:

Ksp of CdF₂ is:

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When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:

6.44x10⁻³ = [X] [2X]²

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X = 0.1172M

<h3>[F⁻] = 0.2344M</h3><h3 />

Ksp of LiF is:

LiF(s) ⇄ Li⁺(aq) + F⁻(aq)

Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]

When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:

1.84x10⁻³ = [X] [X]

1.84x10⁻³ = X²

X = 0.0429

<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
7 0
3 years ago
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