Options are as follow,
A. P₄O₁₀
B. PCl₅
C. Ca₃(PO₄)₂
<span>D. KH</span>₂PO₃
Answer;
The correct option is Option-D (KH₂PO₃).
Explanation:
Oxidation number of Phosphorous is calculated as,
As the normal oxidation number of K is +1, H is +1 and O is -2 so,
KH₂PO₃ = 0
0 shows that the molecule is neutral. Now puttion values of all elemnts except P,
(+1) + (+1)₂ + P + (-2)₃ = 0
+1 + 2 + P - 6 = 0
+3 + P - 6 = 0
+3 + P = +6
P = +6 - 3
P = +3
Molar mass NaOH ( sodium hydroxide) = 40.0 g/mol
Number of moles:
n = mass / molar mass
n = 10.0 / 40.0
n = 0.25 moles
Volume = 2.0 L
Molarity = number of moles / volume in liters
M = 0.25 / 2.0
M = 0.125 M
answer D
hope this helps!
Answer:
Mole fraction of nitrogen = 0.52
Explanation:
Given data:
Temperature = 31.2 °C
Pressure = 870.2 mmHg
Volume = 15.1 L
Mass of mixture = 24.1 g
Mole fraction of nitrogen = ?
Solution:
Pressure conversion:
870.2 /760 = 1.12 atm
Temperature conversion:
31.2 + 273 = 304.2 K
Total number of moles:
PV = nRT
n = PV/RT
n = 1.12 atm × 15.1 L / 0.0821 L.atm. mol⁻¹.K⁻¹ × 304.2 K
n = 16.9 L.atm. /25 L.atm. mol⁻¹
n = 0.676 mol
Number of moles of nitrogen are = x
Then the number of moles of CO₂ = 0.676 - x
Mass of nitrogen = x mol . 28 g/mol and for CO₂ Mass = 44 g/mol ( 0.676 - x)
24.1 = 28x + ( 29.7 -44x)
24.1 - 29.7 = 28x - 44x
-5.6 = -16 x
x = 0.35
Mole fraction of nitrogen:
Mole fraction of nitrogen = moles of nitrogen / total number of moles
Mole fraction of nitrogen = 0.35 mol / 0.676 mol
Mole fraction of nitrogen = 0.52
Answer:
0.33 mol of Fe₂O₃
Explanation:
For the reaction given in the question -
Fe₂O₃ + 5 C + 3O₂ → 2 Fe + 5 CO₂
Balancing the above reaction ,
2Fe₂O₃ + 5 C + 3O₂ → 2 Fe + 5 CO₂
According to the above balanced reaction ,
<u>2 mol of Fe₂O₃ will react with 5 mol of C and 3 mol of O₂ to give 2 mol of Fe and 5 mol of CO₂ .</u>
Hence ,
2 mol of Fe₂O₃ will react with 3 mol of O₂
or ,
3 mol of O₂ ------ 2 mol of Fe₂O₃
Using unitary method ,
1 mol of O₂ ------ 2 / 3 mol of Fe₂O₃
From the question , for 0.49 mol of O₂ ,
0.49 mol of O₂ ------- 2 / 3 * 0.49 mol of Fe₂O₃ = 0.33 mol of Fe₂O₃
