Explanation:
Assuming that moles of nitrogen present are 0.227 and moles of hydrogen are 0.681. And, initially there are 0.908 moles of gas particles.
This means that, for
moles of + moles of = 0.908 mol
Since, 2 moles of = = 0.454 mol
As it is known that the ideal gas equation is PV = nRT
And, as the temperature and volume were kept constant, so we can write
=
=
=
= 5.2 atm
Therefore, we can conclude that the expected pressure after the reaction was completed is 5.2 atm.
Answer:
Rate ≅ 1.01 M/s (3 sig. figs.)
Explanation:
Given A(g) + B(g) => AB(g)
Rate = k[A(g)][B(g)]²
at Rate (1) = 0.239M/s = k[2.00M][2.00M]² => k = (0.239M/s) / (2.00M)(2.00M)²
k = 0.29875 M⁻²·s⁻¹
Rate (2) = k[A(g)][B(g)]² = (0.29875M⁻²·s⁻¹)(4.81M)(2.65M)² = 1.009124472 M/s (calc. ans.) ≅ 1.01 M/s (3 sig. figs.)
Answer:
3 × 10¯¹⁰
Explanation:
9×10² ÷ 3×10¹²
The above expression can be simplified as follow:
9×10² ÷ 3×10¹²
Recall:
9 = 3²
9×10² ÷ 3×10¹² = 3²×10² ÷ 3×10¹²
Recall:
a^m ÷ a^n = a^(m – n)
3²×10² ÷ 3×10¹² = 3^(2 – 1) × 10^(2 – 12)
= 3¹ × 10¯¹⁰
Recall:
a¹ = a
3¹ × 10¯¹⁰ = 3 × 10¯¹⁰
Answer:
by using the chromatography you can separate different color