Answer:
Percent yield of reaction is 75%.
Explanation:
Given data:
Moles of A = 2.0 mol
Moles of B = 4.0 mol
Moles of C formed = 1.8 mol
Percent yield of reaction = ?
Solution:
Chemical equation:
2A + 5B → 3C + D
We will compare the moles of A and B with C from balance chemical equation.
A : C
2 : 3
B : C
5 : 3
4 : 3/5×4 = 2.4 mol
Number of moles of C formed by 4 moles B are less thus B is limiting reactant and will limit the yield of C.
Percent yield:
Percent yield = actual yield /theoretical yield × 100
Percent yield = 1.8 mol / 2.4 × 100
Percent yield =75%
Answer:
D.
Explanation:
Only 0.0035% of the electromagnetic spectrum is visible to the human eye
Hopefully this helped :)
Answer:
Explanation:
<u>1. Word equation:</u>
- <em>mercury(II) oxide → mercury + oxygen </em>
<u>2. Balanced molecular equation:</u>
<u>3. Mole ratio</u>
Write the ratio of the coefficients of the substances that are object of the problem:
<u>4. Calculate the number of moles of O₂(g)</u>
Use the equation for ideal gases:
<u>5. Calculate the number of moles of HgO</u>
<u>6. Convert to mass</u>
- mass = # moles × molar mass
- molar mass of HgO: 216.591g/mol
- mass = 0.315mol × 216.591g/mol = 68.3g
The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
Answer:
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
Explanation:
Step 1: Data given
Molar mass helium = 4.0 g/mol
Molar mass O2 = 32 g/mol
Step 2: Graham's law
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the molecular mass : 1/(Mr)^0.5
Rate of escape for He = 1/(4.0)^0.5 = 0.5
Rate of escape for O2 = 1/(32)^0.5 = 0.177
The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s
