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Answer:
C
Explanation:
Fresh water fish need freshwater.
We know, 1 m³ of space can hold 1000 l of the substance.
⇛ 1 m³ = 1000 l----(1)
And, 1 l is 1000 times more than 1 ml
⇛ 1 l = 1000 ml------(2)
So, From (1) and (2),
⇛ 1 m³ = 1000 × 1000 ml
⇛ 1m³ = 1000000 ml
We had to find,
⇛ 1.40 m³ = 1.40 × 1000000 ml
⇛ 1.40 m³ = 140/100 × 1000000 ml
⇛ 1.40 m³ = 1400000 ml
⇛ 1.40 m³ = 14,00,000 ml / 14 × 10⁵ ml / 1.4 × 10⁶ ml
☃️ <u>So</u><u>,</u><u> </u><u>1.40</u><u> </u><u>m</u><u>³</u><u> </u><u>=</u><u> </u><u>1</u><u>4</u><u> </u><u>×</u><u> </u><u>1</u><u>0</u><u>⁵</u><u> </u><u>m</u><u>l</u><u> </u><u>/</u><u> </u><u>1.4</u><u> </u><u>×</u><u> </u><u>10</u><u>⁶</u><u> </u><u>ml</u><u>.</u>
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1) Write the balanced chemical equation
4NH3 + 5O2 ----> 4NO + 6H2O
2) Write the molar ratios
4 mol NH3 : 5 mol O2 : 4 mol NO : 6 mol H2O
3) Write the proportions with the desired quantity of NO and the unknown quantity of O2
5 mol O2 / 4 mol NO = x / 18.0 mol NO
Solve for x: x = 18.0 mol NO * 5 mol O2 / 4 mol NO = 22.5 mol O2
4) Use the ideal gas equation to convert moles to volume
pV = nRT => V = nRT / p
n = 22.5 mol
R = 0.082 atm*liter / K * mol
T = 33 + 273.15 = 306.15 K
p = 874/760 atm = 1.15 atm
V = 22.5 mol * 0.082 atm*liter/K*mol * 306.15 K / 1.15 atm = 491.17 liter
Answer: 491 liters of O2
