The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.
<h3>How do we calculate the partial pressure of gas?</h3>
Partial pressure of particular gas will be calculated as:
p = nP, where
- P = total pressure = 748 mmHg
- n is the mole fraction which can be calculated as:
- n = moles of gas / total moles of gas
Moles will be calculated as:
- n = W/M, where
- W = given mass
- M = molar mass
Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole
Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole
Mole fraction of hydrogen = 1 / (1+0.5) = 0.6
Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm
Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.
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Answer:
To calculate the number of atoms in a sample, divide its weight in grams by the amu atomic mass from the periodic table, then multiply the result by Avogadro's number: 6.02 x 10^23. Set up Equation Express the relationship of the three pieces of information you need to calculate the number of atoms in the sample in the form of an equation.
Muscles bones no on organs and no on skin
Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Answer:
A. co dominace
incomplete dominance is when both genes show, this would result in the chicken being a black and white color
Explanation:
