Answer:
Water heaters must have vent to the outside of the heater to remove exhaust air from the water heater, the vent go directly through it walls. It is likened to the vent in natural gas water heater which have a vent to the outside that remove by products of combustion and can also bring in air combustion.
The edifice from the hot water vent are made of metals or plastic materials depending on the venting systems.
The way to working out the numbers is to increase the measure of HNO3 required by the molarity to discover what number of moles you require: 0.115. You ought to have the capacity to make sense of the recipe weight H is 1, N is 14, O is 16. The result of the quantity of moles duplicated by the recipe weight ought to give an esteem in grams. You can utilize the thickness to change over to a volume of HNO3 to add to the right volume of water.
Answer:
0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl
Explanation:
This is the reaction:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)
To make 3 moles of H₂, we need 2 moles of Al.
By conditions given, we will find out how many moles of H₂ do we have.
Let's use the Ideal Gas Law
P. V = n . R . T
1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K
(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n
0.182 mol = n
So the rule of three will be:
If 3 moles of H₂ came from 2 moles of Al
0.182 moles of H₂ will come from x
(0.182 .2) / 3 = 0.121 moles
10 miles per hour because 500 divided by 50 is 10
Answer:
Explanation:
Question 7.
We can use the Combined Gas Laws to solve this question.
a) Data
p₁ = 1.88 atm; p₂ = 2.50 atm
V₁ = 285 mL; V₂ = 435 mL
T₁ = 355 K; T₂ = ?
b) Calculation
Question 8. I
We can use the Ideal Gas Law to solve this question.
pV = nRT
n = m/M
pV = (m/M)RT = mRT/M
a) Data:
p = 4.58 atm
V = 13.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 385 K
M = 46.01 g/mol
(b) Calculation
