Answer:
The final pressure of the gas is:- 21.3 kPa
Explanation:
Using Boyle's law

Given ,
V₁ = 10.0 L
V₂ = 45.0 L
P₁ = 96.0 kPa
P₂ = ?
Using above equation as:




The final pressure of the gas is:- 21.3 kPa
Answer:
atoms of hydrogen are there in
35.0 grams of hydrogen gas.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
of particles.
To calculate the moles, we use the equation:
1 mole of hydrogen
=
atoms
17.5 mole of hydrogen
=
atoms
There are
atoms of hydrogen are there in
35.0 grams of hydrogen gas.
Answer:
moles H₂O = 10
Explanation:
The mass of Na₂CO₃⋅xH₂O is 3.837 g and the mass of Na₂CO₃ is 1.42g
Therefore the mass of xH₂O is 3.837 - 1.42 = 2.417 g
The molar mass of Na₂CO₃ is 106 g/mol and for H₂O is 18 g/mol
The moles of Na₂CO₃ and H₂O in the sample are:
Na₂CO₃ = 1.42 / 106 = 0.01340 moles
H₂O = 2.417 / 18 = 0.1343
Now using rule of three :
1 mole of Na₂CO₃ has x moles of H₂O
0.01340 moles of Na₂CO₃ has 0.1343 moles of H₂O
x = 1 * 0.1343 / 0.01340 = 10
Answer:
6
Explanation:
p orbital can hold up to six electron. Argon electron configuration will be 1s²2s²2p⁶3s²3p⁶