How many seconds are required to produce 4.00 g of aluminum metal from the electrolysis of molten alcl3 with an electrical curre

The chemical equation can be expressed as:

2H2 + O2 = 2H2O

Given the amounts of the reactants, we need to identify the limiting reactant before calculating the amount of product generated.

4.0 g H2 ( 1 mol / 2.02 g ) = 1.98 mol H2
5.0 g O2 ( 1 mol / 32 g ) = 0.1563 mol O2

The limiting reactant is O2, as it will be fully consumed in the reaction.

0.1563 mol O2 ( 2 mol H2O / 1 mol O2 ) ( 18.02 g / mol ) = 5.6 g H2O will be produced

Answer: The mole fraction of nitrogen is calculated to be 0.4615.

Explanation: In the mixture of nitrogen () and hydrogen (), the mole ratio established is 1: 1.5.

At this juncture, we identify that () serves as the limiting reagent.

When 0.4 moles of () are produced, it will require 0.4 moles of () along with 3.4 moles of ().

The total amount of remaining () ends up being 0.6, while the amount of () left in the mixture is 0.3 moles.

The mole fraction of () is computed as 0.6 divided by the sum of 0.6, 0.4, and 0.3, resulting in 0.4615.

Answer:

The alteration in the state of matter occurs when sufficient thermal energy is either added to or taken away from an object.

The updated volume of the balloon, when cooled at a constant pressure, is 3.98 L

Explanation

This new volume was determined using the formula from Charles' law

presented as V1/ T1 = V2/T2 where,

V1 = 4.24 L

T1= 23°C converted to kelvin = 23 + 273 = 296 K

T2 = 5.00°C converted to kelvin = 5.00 + 273 = 278 K

V2 =?

By rearranging the equation to isolate V2, we can find it by multiplying both sides by T2

V2 = V1 x T2/ T1

This results in V2 = (4.24 L x 278 K) / 296 K = 3.98 L