Answer:
q1 = mCpΔT
= 18.016g × 1.84J/g.K × (418.15-373.15)
= 1491.72 J
q2 = n×ΔH vap
= 1mol ×44.0kJ/mol
= 44KJ
∴ qtotal = q1+ q2
= 1.498kJ + 44.0kJ
= 45.498KJ
Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.
Because society has shaped our mind on what we shall think about things
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
Answer:
0.52 mol
Explanation:
Using the general gas equation formula:
PV = nRT
Where;
P = pressure (atm)
V = volume (Liters)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
At STP (standard temperature and pressure), temperature of a gas is 273K, while its pressure is 1 atm
Using PV = nRT
n = PV/RT
n = (1 × 11.74) ÷ (0.0821 × 273)
n = 11.74 ÷ 22.41
n = 0.52 mol
There are 0.52 moles in the basketball
