Answer:
The molecular weight of CO is 28.01g/mole.
We can use the heat equation,
Q = mcΔT
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = <span>145 g
</span>c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹
Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.
C=10⁻⁶ mol/L
pH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg10⁻⁶=14-6=8
B. pH = 8
Answer : The mass of 
needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of 
.
Now we have to calculate the moles of .
The balanced chemical reaction will be,
![PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4](https://tex.z-dn.net/?f=PbSO_4%2BNa_2CrO_4%5Crightarrow%20PbCrO_4%2BNa_2SO_4%5Btex%5D%3C%2Fp%3E%3Cp%3EFrom%20the%20balanced%20chemical%20reaction%2C%20we%20conclude%20that%3C%2Fp%3E%3Cp%3EAs%2C%201%20mole%20of%20%5Btex%5DPbCrO_4)
produced from 1 mole of
So, 0.005 mole of produced from 0.005 mole of
Now we have to calculate the mass of
Therefore, the mass of needed are, 1.515 grams.
Answer:
V = 22.41 L
Explanation:
Given data:
Mass of nitrogen = 14.0 g
Volume of gas at STP = ?
Gas constant = 0.0821 atm.L/mol.K
Solution:
Number of moles of gas:
Number of moles = mass/molar mass
Number of moles= 14 g/ 14 g/mol
Number of moles = 1 mol
Volume of gas:
PV = nRT
1 atm × V = 1 mol × 0.0821 atm.L/mol.K × 273 K
V = 22.41 atm.L / 1 atm
V = 22.41 L
