Missing question: what is the density of 53.4 wt% aqueous NaOH if 16.7 mL of the solution diluted to 2.00L gives 0.169 M NaOH?
Answer is: density is 1.52 g/mL.
c₁(NaOH) = ?; molarity of concentrated sodium hydroxide.
V₁(NaOH) = 16.7 mL; volume of concentrated sodium hydroxide.
c₂(NaOH) = 0.169 M; molarity of diluted sodium hydroxide.
V₂(NaOH) = 2.00 L · 1000 mL/L = 2000 mL; volume of diluted sodium hydroxide.
Use equation: c₁V₁ = c₂V₂.
c₁ = c₂V₂ / V₁.
c₁ = 0.169 M · 2000 mL / 16.7 mL.
c₁(NaOH) = 20.23 M.
m(NaOH) = 20.23 mol · 40 g/ml.
m(NaOH) = 809.53 g.
The mass fraction is the ratio of one substance (in this example sodium hydroxide) with mass to the mass of the total mixture (solution).
Make proportion: m(NaOH) : m(solution) = 53.4 g : 100 g.
m(solution) = 1516 g in one liter of solution.
d(solution) = 1516 g/L = 1.52 g/mL.
Answer:
The amount of ammonia needed is 33.3 g
Therefore, we can also say 1.96 moles of NH₃
Explanation:
The reaction is: 3CuO(s) + 2NH₃(g) → 3H₂O(l) + 3Cu(s) + N₂(g)
If we see stoichiometry, 3 moles of water can be produced by 2 moles of NH₃. We propose this rule of three:
3 moles of water can be produced by 2 mole of ammonia
Then, 2.94 moles of water, must be produced by (2.94 . 2) /3 =1.96 moles of NH₃
If we convert the moles to mass. 1.96 mol . 17 g /1mol = 33.3 g
Explanation:
Given parameters:
Mass of sample = 235g
Molecular mass of sample = 128.1g
Empirical formula = CH₂O
Unknown:
Mass of each element in the sample = ?
Solution:
To solve this problem, we must know that the empirical formula of any compound is the simplest ratio of the atoms it contains. This is not the true formula of the compound.
Molecular formula = (Empirical formula)ₙ
Let us find the molecular mass of the sample;
CH₂O = 12 + 2(1) + 16 = 30g
128.1 = (30)n
n = 4
The molecular formula of the compound is; (CH₂O)₄ = C₄H₈O₄
Now to find the grams of each element in the sample;
Express the molecular mass of each element and that of the compound as a fraction and multiply with the given mass;
For C;
= 88.06g
H; 
= 14.68g
O: 
= 117.41g
learn more:
Mass composition brainly.com/question/3018544
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Answer:
This is the conversion factor, we have to use:
44 g / 1mol
Explanation:
The reaction for the methane combustion is:
CH₄ + 2O₂ → CO₂ + 2H₂O
First of all we use this conversion factor to determine the moles of methane, we used.
50.6 g . 1 mol / 16 g = 3.16 mol
So ratio is 1:1, then 3.16 mol of methane will produce 3.16 moles of CO₂
To calculate the grams of produced dioxide:
3.16 mol . 44 g / 1mol = 139.04 g
