I think the answer is 101.2 L
<span>The products of the light-dependent reactions are used to help 'fuel' the light-independent reactions.
</span><span>Example:
NADPH and ATP are produced during the light-dependent reaction for use in the light-independent reaction (the Calvin Cycle). </span>
<span>Answer:
A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound?
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Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different.
25.305% C/12 = 2.108
74.695% Cl/35.5 = 2.104
So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.)
0.044 grams/10 ml = x/22.4 liters
0.044g/0.010 liters = x/22.4 liters
22.4 liters/0.010 liters = 2240 (ratio)
2240 x .044 = 98.56 (actual atomic weight)
CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole.
This is sufficiient to distinguish C2CL2, (dichloroacetylene)
from C6CL6 (hexachlorobenzene) which would
mass 3 times as much.</span>
Using v1/t1=v2/t2
v1=500
v2=?
t1=75=368k
t2=225=498
500/368=v2/498
1.4x498=v2
v2=697.2ml
Answer:
= 67.79 g
Explanation:
The equation for the reaction is;
4Cr(s)+3O2(g)→2Cr2O3(s)
The mass of O2 is 21.4 g, therefore, we find the number of moles of O2;
moles O2 = 21.4 g / 32 g/mol
=0.669 moles
Using mole ratio, we get the moles of Cr2O3;
moles Cr2O3 = 0.669 x 2/3
=0.446 moles
but molar mass of Cr2O3 is 151.99 g/mol
Hence,
The mass Cr2O3 = 0.446 mol x 151.99 g/mol
<u> = 67.79 g
</u>
