Answer:
1528.3L
Explanation:
To solve this problem we should know this formula:
V₁ / T₁ = V₂ / T₂
We must convert the values of T° to Absolute T° (T° in K)
21°C + 273 = 294K
70°C + 273 = 343K
Now we can replace the data
1310L / 294K = V₂ / 343K
V₂ = (1310L / 294K) . 343K → 1528.3L
If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too
Answer:
For these problems, we need to compare the theoretical yield that we'd get from performing stoichiometry to the actual yield stated in the problem. % yield is the actual yield/theoretical yield x 100%
Cu + 2 AgNO₠→ Cu(NOâ‚)â‚‚ + 2 Ag ==> each mole of copper yields two moles of silver
12.7-g Cu x ( 1 mol Cu /63.5-g Cu) x ( 2 mol Ag / 1 mol Cu) x (108-g Ag / 1 mol Ag) = 43.2-g Ag. This is the theoretical yield. Now, since we got 38.1-g Ag our % yield is:
38.1-g/43.2-g x 100% = 88.2%
Explanation:
The answer is: 0.158 mol
You find this by doing:
number of moles (n) = mass (m) / molar mass (M)
n=158.034/25.0
The balanced chemical reaction is:
<span>C4H8O2 + 2H2--> 2C2H6O
</span>A. How many grams of ethyl alcohol are produced by reaction of 2.7 mol of ethyl acetate with H2? 2.7 mol C4H8O2 ( 2 mol C2H6O / 1 mol C4H8O2) (46.07 g / 1 mol) = 248.78 g ethyl alcohol
B. How many grams of ethyl alcohol are produced by reaction of 13.0g of ethyl acetate with H2?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol C2H6O / 1 mol C4H8O2) (<span>46.07 g / 1 mol) = 13.59 g ethyl alcohol</span>
C. How many grams of H2 are needed to react with 13.0g of ethyl acetate?
13.0 g C4H8O2 ( 1 mol / 88.11 g) ( 2 mol H2 / 1 mol C4H8O2) (2.02 g / 1 mol) = 0.5961 g H2
Answer:
0%
Explanation:
- The probability that both of the parents are homozygous recessive is zero.
- In our case, The gene for tallness is dominant over the gene for dwarfism
- Therefore; TT is homozygous dominant while tt is homozygous recessive and Tt is heterozygous.
- Heterozygous genotype (Tt) will exhibit a tall phenotype.
- When two parents are crossed and all the offsprings are tall then the possible genotype of the parents is either;
- Homozygous dominant for both parents or
TT x TT
2. One parent is homozygous dominant and the other is homozygous recessive.
TT x Tt
- Therefore, the probability of having both parents as homozygous recessive when all the offspring are tall is Zero.
