The number of moles of the magnesium (mg) is 0.00067 mol.
The number of moles of hydrogen gas is 0.0008 mol.
The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.
<h3>
Number of moles of the magnesium (mg)</h3>
The number of moles of the magnesium (mg) is calculated as follows;
number of moles = reacting mass / molar mass
molar mass of magnesium (mg) = 24 g/mol
number of moles = 0.016 g / 24 g/mol = 0.00067 mol.
<h3>Number of moles of hydrogen gas</h3>
PV = nRT
n = PV/RT
Apply Boyle's law to determine the change in volume.
P1V1 = P2V2
V2 = (P1V1)/P2
V2 = (101.39 x 146)/(116.54)
V2 = 127.02 mL
Now determine the number of moles using the following value of ideal constant.
R = 8.314 LkPa/mol.K
n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)
n = 0.0008
<h3>Volume of 1 mole of hydrogen gas at STP</h3>
V = nRT/P
V = (1 x 8.314 x 273) / (101.325)
V = 22.4 L
Learn more about number of moles here: brainly.com/question/13314627
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<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
<u>Explanation:</u>
To calculate the mole percent of a substance, we use the equation:

Mass percent means that the mass of a substance is present in 100 grams of mixture
To calculate the number of moles, we use the equation:

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively
Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.

where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.

It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.

The oxidation state of Pb in
is 2. So, moles deposited by Pb is as follows.

It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.

Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.
when iron and oxygen the reactin form Fe2O3 balance equation for this is <span>4Fe + 3O2---> 2Fe2O3
9moles of O2 are needed to produce 6mol of Fe2O3 since the ratio of oxygen to iron(II)oxide is 3:2
hope its help</span>