Answer: (1). There are 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.
(2). The density of gaseous arsine is 3.45 g/L.
Explanation:
1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
2). As number of moles are also equal to mass of a substance divided by its molar mass.
So, number of moles of Arsine 
(molar mass = 77.95 g/mol) is as follows.
Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.
Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.
If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
Answer:
Volume of acid in the solution, 53.87 mL
Explanation:
This can be solved with a simple formula
(Volume of solution which is acid / Total volume) . 100 = % acid
(Volume of solution which is acid / 341 mL) . 100 = 15.8
Volume of solution which is acid / 341 mL = 0.158
Volume of solution which is acid = 0.158 . 341 mL
Volume of solution which is acid = 53.87 mL
Answer:
0
Explanation:
pure elements will always possess an oxidation number of 0, regardless of their charge.
Answer:
Arrow A:
✔ 74.6 kJ
Arrow B:
✔ exothermic
Arrow C:
✔ has a magnitude that is less than that of B
Arrow D:
✔ represents the overall enthalpy of reaction
Explanation:
