N2 + 3H2 = 2NH3
in this question, we are dealing with only NH3 and H2 so we only focus on that
since the ratio of H2 to 2NH3 is 3:2, we say that
3 liters of H2 = 2 liters of 2NH3
3.6 litres of H2 = x liters of 2NH3
We cross multiple to give:
3 × x = 3.6 × 2
3x = 7.2
Divide both sides by 3
x = 7.2 ÷ 3
x = 2.4liters
Answer:
0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar
<span>Answer: 8.15s
</span><span />
<span>Explanation:
</span><span />
<span>1) A first order reaction is that whose rate is proportional to the concenration of the reactant:
</span><span />
<span>r = k [N]
</span><span />
<span>r = - d[N]/dt =
</span><span />
<span>=> -d[N]/dt = k [N]
</span><span />
<span>2) When you integrate you get:
</span><span />
<span>N - No = - kt
</span>
<span></span><span /><span>
3) Half life => N = No / 2, t = t'
</span><span />
<span>=> No - No/ 2 = kt' => No /2 = kt' => t' = (No/2) / k
</span><span />
<span>3) Plug in the data given: No = 0.884M, and k = 5.42x10⁻²M/s
</span>
<span /><span /><span>
t' = (0.884M/2) / (5.42x10⁻²M/s) = 8.15s</span>
Answer:
C11H25SO4
Explanation:
The total mass of the compound is 253.4 g, so, the mass of each element will be:
C: 52.14% of 253.4 = 0.5214x253.4 = 132.12 g
H: 9.946% of 253.4 = 0.09946x253.4 = 25.20 g
S: 12.66% of 253.4 = 0.1266x253.4 = 32.08 g
O: 25.26% of 253.4 = 0.2526x253.4 = 64.00 g
The molar mass are: C = 12 g/mol, H 1 g/mol, S = 32 g/mol, and O = 16 g/mol
So, to know how much moles will be, just divide the mass calculated above for the molar mass:
C: 132.12/12 = 11 moles
H: 25.20/ 1 = 25 moles
S: 32.08/32 = 1 mol
O: 64.00/16 = 4 moles
So the molecular formula is C11H25SO4
