Answer:
The answer to your question is: 1538095.2 kg of NH3
Explanation:
MW HNO3 = 63 kg
MW NO2 = 46 kg
3 NO2(g) + H2O(l)--- 2 HNO3(aq) + NO(g)
3(46) kg-------------- 2(63) kg
x --------------- 7600000 kg
x = 7600000 x 138/126 = 8323809.5 kg og NO2
MW NO = 30
2 NO(g) + O2(g)---2 NO2(g)
2(30) ------------------2(46)
x ---------------- 8323809.5 kg
x = 8323809.5 x 60/92 = 5428571.4 kg of NO
MW NH3 = 17 kg
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)
4(17) -------------------- 4(30)
x ----------------------- 5428571.4
x = 5428571.4 x 34 / 120
x = 1538095.2 kg of NH3
<span>0.70 mol/0.250 L = 2.8 M</span>
To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns.
For the first equation, we do a mass balance:
mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar
Assuming they have the same densities, then we can write this equation in terms of volume.
V(100%) + V(13%) = V(42%)
we let x = V(100%)
y = V(13%)
x + y = 150
For the second equation, we do a component balance:
1.00x + .13y = 150(.42)
x + .13y = 63
The two equations are
x + y = 150
x + .13y = 63
Solving for x and y,
x = 50
y = 100
Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.
Answer:
C. The more niches in an ecosystem, the greater the biodiversity.
Explanation:
Dilution of the solution can be calculated by the formula of the molarity and volume. The initial volume of 2.50 M solution was 30 mL.
<h3>What is the relationship between molar concentration and dilution?</h3>
Molar concentration or the dilution factor is in an inverse relationship and with an increase in the dilution, the molarity of the solution decreases.
Given,
Initial molarity = 2.50 M
initial volume = ?
Final molarity = 0.750 M
Final volume = 100.0 ml
Substituting values in the formula:
Therefore, 30 mL was the initial volume of the solution before it was diluted.
Learn more about dilution here:
brainly.com/question/26896011
