Answer:
the volume occupied by 3.0 g of the gas is 16.8 L.
Explanation:
Given;
initial reacting mass of the helium gas, m₁ = 4.0 g
volume occupied by the helium gas, V = 22.4 L
pressure of the gas, P = 1 .0 atm
temperature of the gas, T = 0⁰C = 273 K
atomic mass of helium gas, M = 4.0 g/mol
initial number of moles of the gas is calculated as follows;
The number of moles of the gas when the reacting mass is 3.0 g;
m₂ = 3.0 g
The volume of the gas at 0.75 mol is determined using ideal gas law;
PV = nRT
Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.
Answer:
A molecule is a group of two or more atoms held together by chemical bonds. A compound is a substance which is formed by two or more different types of elements which are united chemically in a fixed proportion. All molecules are not compounds.
Good luck !
Using the law of dilution :
Mi x Vi = Mf x Vf
2.00 x Vi = 0.15 x 100.0
2.00 x Vi = 15
Vi = 15 / 2.00
Vi = 7.5 mL
hope this helps!
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
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