1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
Answer:
69.8 kilo Pasacl is the pressure of the hydrogen gas.
Explanation:

Pressure at which hydrogen gas collected = p = 101.2 kilo Pascals
Vapor pressure water =
= 31.4 kilo Pascals
The pressure of hydrogen gas = P
The pressure at which gas was collected was sum of vapor pressure of water and hydrogen gas.


69.8 kilo Pasacl is the pressure of the hydrogen gas.
Mercury is a homogenous mixture of two or more pure substances
<span>Use the van't Hoff equation:
ln
(
K2
K1
)
=
Δ
HÂş
R
(
1
T1
â’
1
T2
)
ln
(
K2
7.6*10^-3
)
=
-14,200 J
8.314
(
1
298
â’
1
333
)
ln
(
K2
7.6*10^-3
)
=
â’
1708
(
0.00035
)
ln
(
K2
0.0076
)
=
â’
0.598
Apply log rule
a
=
log
b
b
a
-0.598 =
ln
(
e
â’
0.598
)
=
ln
(
1
e
0.598
)
Multiply both sides with e^0.598
K
2
e
0.598
= 0.0076
K
e
0.598
e
0.598
=
0.0076
e
0.598
K
2
=
0.0076
e
0.598
=
4.2
â‹…
10
â’
3
K2
=
4.2
â‹…
10
â’
3</span>
Answer:
well try talking to him first and being his friend. introduce yourself and try not to be too awkward or shy. can't reltate lol. after that if he really likes you, he'll initiate stuffs ^^
Explanation: