The question is incomplete, here is the complete question:
The solubility of substance X in water at a certain temperature is 35.7 g /100. g. Suppose that you have 330.0 g of substance X. What is the minimum volume of water you would need to dissolve it all? (Assume that the density of water is 1.00 g/mL.)
<u>Answer:</u> The minimum volume of water that would be needed is 940.17 mL
<u>Explanation:</u>
We are given:
Solubility of substance X in water = 35.1 g/100 g
This means that 35.1 grams of substance X is dissolved in 100 grams of water
Applying unitary method:
If 35.1 grams of substance X is dissolved in 100 grams of water
So, 330.0 grams of substance X will be dissolved in = 
of water
To calculate the volume of water, we use the equation:
Density of water = 1 g/mL
Mass of water = 940.17 g
Putting values in above equation, we get:
Hence, the minimum volume of water that would be needed is 940.17 mL
Answer:
10%
Explanation:
Given data:
Actual yield = 0.13 moles
Percent yield = ?
Solution:
Balanced chemical equation.
3H₂ + N₂ → 2NH₃
First of all we will calculate the mass of given moles of ammonia.
Number of moles = mass / molar mass
Mass = 0.13 × 17 g/mol
Mass = 2.21 g
2.21 g is experimental yield of ammonia.
Hydrogen is limiting reactant because only two moles of hydrogen present.
we will compare the moles of hydrogen and ammonia from balance chemical equation.
H₂ : NH₃
3 : 2
2 : 2/3×2 = 1.33 moles
Mass of ammonia
Mass = number of moles × molar mass
Mass = 1.33 mol × 17 g/mol
Mass = 22.61 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 2.21 g / 22.61 g
Percent yield =9.8% which is almost 10%
Answer:
1.35
Explanation Use the mole formla and multiply and them apply the sig fig rules and you will get the moles of water hope this helps god bless
Nitrous acid<span> dissociates as follows:
</span>
HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq)
According to the equation, an acid constant has the following form:
Ka = [H⁺] × [NO₂⁻ ] / [HNO₂]
From pH, we can calculate the concentration of H⁺ and NO₂⁻:
[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]
Now, the acid constant can be calculated:
Ka = 0.00234 x 0.00234 / 0.015 = 3.66 x 10⁻⁴
And finally,
pKa = -log Ka = 3.44
