Answer:
The percent yield of the reaction is 35 %
Explanation:
In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.
Let's verify the moles that were used in the reaction.
2.05 g . 1mol/ 32 g = 0.0640 mol
In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.
Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).
1atm . 0.550L = n . 0.082 . 295K
(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles
Percent yield of reaction = (Real yield / Theoretical yield) . 100
(0.0225 / 0.0640) . 100 = 35%
Natural abundance of oxygen I think
Answer:
Yes, Pb3(PO4)2.
Explanation:
Hello there!
In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.
Regards!
Answer:
Q = 2647 J
Explanation:
Specific heat capacity is the amount of energy required by one Kg of a substance to raise its temperature by 1 °C.
In thermodynamics the equation used is as follow,
Q = m Cp ΔT
Where;
Q = Heat = ?
m = mass = 660 g
Cp = Specific Heat Capacity = 0.3850 J.g⁻¹.°C⁻¹
ΔT = Change in Temperature = 23.35 °C - 12.93 °C = 10.42 °C
Putting values in eq. 1,
Q = 660 g × 0.3850 J.g⁻¹.°C⁻¹ × 10.42 °C
Q = 2647 J
hey mate here is ur answer
solution
mass{m}=3 gram
=3/1000
volume{v}=16cm
=16/100
density=m/v
=3/1000÷16/100
=3/160
=0.01875kg/m3
