Answer:
16.46 g.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Cu + Zn(NO₃)₂ → Zn + 2Cu(NO₃).</em>
- It is clear that 2.0 moles of Cu reacts with 1.0 mole of Zn(NO₃)₂ to produce 1.0 mole of Zn and 2.0 moles of Cu(NO₃).
- We need to calculate the number of moles of the reacted Cu (32.0 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Cu = mass / atomic mass = (32.0 g) / (63.546 g/mol) = 0.503 mol.
<u><em>Using cross multiplication:</em></u>
2.0 moles of Cu produces → 1.0 mole of Zn, from the stichiometry.
0.503 mole of Cu produces → ??? mole of Zn.
- The no. of moles of Zn produced = (1.0 mol)(0.503 mol) / (2.0 mol) = 0.2517 mol.
∴ The grams of Zn produced = no. of moles x atomic mass of Zn = (0.2517 mol)(65.38 g/mol) = 16.46 g.
Data:
Molar Mass of NaOH = 40 g/mol
Solving: <span>According to the Law Avogradro, we have in 1 mole of a substance, 6.02x10²³ atoms/mol or molecules
</span>
1 mol -------------------- 6.02*10²³ molecules
y mol -------------------- 2.70*10²² molecules
6.02*10²³y = 0.270*10²³
Solving: <span>Find the mass value now
</span>
40 g ----------------- 1 mol of NaOH
x g ------------- 0.04 mol of NaOH
Answer:
The mass is 1.6 grams
Answer:
so 0.15 moles X 22.4 dm3/mole=3.36 dm3. Next we find the moles of hexane combusted, and then the moles of CO2. Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm3.
Answer:
The first row of elements fits in period <u>6</u>, after the element <u>lanthanum (La)</u>. The second row of elements fits in period <u>7</u>, after the element <u>actinium (Ac). </u>
I hope this helps!
The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
<h3>How to calculate empirical formula?</h3>
The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
The empirical formula of the given compound can be calculated as follows:
- Hafnium = 55.7% = 55.7g
- Chlorine = 44.3% = 44.3g
First, we convert mass values to moles by dividing by the molar mass of each element
- Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
- Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol
Next, we divide each mole value by the smallest
- Hafnium = 0.312 ÷ 0.312 = 1
- Chlorine = 1.25 ÷ 0.312 = 4
Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
Learn more about empirical formula at: brainly.com/question/14044066
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