Yes because molecules is solid
Answer:
CH₂ ; 67.1 %
Explanation:
To determine the empirical formula we need to find what the mole ratio is in whole numbers of the atoms in the compound. To do that we will first need the atomic weights of C and H and then perform our calculation
Assume 100 grams of the compound.
# mol C = 85.7 g / 12.01 g/mol = 7.14 mol
# mol H = 14.3 g / 1.008 g/mol = 14.19 mol
The proportion is 14.9 mol H/ 7.14 mol C = 2 mol H/ 1 mol C
So the empirical formula is CH₂
For the second part we will need to first calculate the theoretical yield for the 12.03 g NaBH₄ reacted and then calculate the percent yield given the 0.295 g B₂H₆ produced.
We need to calculate the moles of NaBH₄ ( M.W = 37.83 g/mol )
1.203 g NaBH₄ / 37.83 g/mol = 0.0318 mol
Theoretical yield from balanced chemical equation:
0.0318 mol NaBH₄ x 1 mol B₂H₆ / mol NaBH₄ = 0.0159 mol B₂H₆
Theoretical mass yield B₂H₆ = 0.0159 mol x 27.66 g/ mol = 0.440 g
% yield = 0.295 g/ 0.440 g x 100 = 67.1 %
Answer:
pure water will have the lowest boiling point.
Missing question: Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of ammonium carbonate and cobalt(II) bromide are combined.<span>Balanced chemical reaction:
(NH</span>₄)₂CO₃(aq) + CoBr₂(aq) → CoCO₃(s) + 2NH₄Br(aq).
Net ionic reaction:
2NH₄⁺(aq) + CO²⁻(aq) + Co²⁺(aq) + 2Br⁻(aq) → CoCO₃ + 2NH₄(aq)+ 2Br(aq).
or CO²⁻(aq) + Co²⁺(aq) → CoCO₃(s).
1.
V = 200 mL (volume)
c = 3 M = 3 mol/L (concentration)
First we convert mL to L:
200 mL = 0.2 L
Then we calculate the moles using the formula: n = V × c = 0.2 L × 3 mol = 0.6 mol
Finally, we just use the molar mass of CaF2 to calculate the actual mass:
molar mass = 78 g/mol
The formula is: m = n × mm (mass = moles × molar mass)
m = 0.6 mol × 78 g/mol = 46.8 g
2.
For this question the steps are exactly like the first question.
V = 50mL = 0.05 L
c = 12 M = 12 mol/L
n = V × c = 0.05 L × 12 mol/L = 0.6 mol
molar mass (HCl) = 36.5 g/mol
m = n × mm = 0.6 mol × 36.5 g/mol = 21.9 g.
3.
The steps for this question are the opposite way.
m(K2CO3) = 250 g
molar mass = 138 g/mol
n = m ÷ mm = 1.81 mol
c = 2 mol/L
V = n ÷ c = 1.81 mol ÷ 2 mol/L = 0.905 L = 905 mL
