Answer:
The concentration of helium in the water is 2.405×10^-4 M
Explanation:
Concentration = Henry's law constant × partial pressure of helium
Henry's law constant = 3.7×10^-4 M/atm
Partial pressure of helium = 0.65 atm
Concentration = 3.7×10^-4 × 0.65 = 2.405×10^-4 M
The larger the kinetic energy of the vehicle, the larger the amount of energy will be needed to stop the vehicle, meaning that faster vehicles have a larger stopping distance
Interesting problem. Thanks for posting.
C2H2 + (3/2)02 ====> H2O + 2CO2
CH4 + 2O2 =====> 2H2O + CO2
The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16
The number of moles of C2H2 = x
The number of moles of CH4 = y
26x + 16y = 230.9 grams
For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18
x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water.
Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2 we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2
Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y = 972.7
Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7
I'm not going to go through the math unless you request me to do so.
x = 8.03 moles
y = 1.38 moles
The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
Answer:
it would be a covalent bond
Explanation:
carbon has 4 valence electrons, while chlorine has 7.
Answer:
The answer to your question is 92.7%
Explanation:
Balanced Chemical reaction
3 Zn + Fe₂(SO₄)₃ ⇒ 2Fe + 3ZnSO₄
Molecular weight
Zinc = 65.4 x 3 = 196.2g
Iron (III) = 56 x 2 = 112 g
Proportions
196.2 g of Zinc ------------------ 112 g of Iron
20.4 g of Zinc ----------------- x
x = (20.4 x 112) / 196.2
x = 2284.8/196.2
x = 11.65 g of Iron
% yield = 
% yield = 0.927 x 100
% yield = 92.7
