Answer:
Explanation:
given that
mass of Ba(NO3)2 = 1.40g
mass of NH2SO3H = 2.50 g
1)to determine the mole of Ba(NO3)2
2) to determine the mass of all three product formed in the reaction
reaction
Ba(NO3)2 + 2NH2SO3H → Ba(NH2SO3)2 + 2HNO3
<u>Solution</u>
we calculate the molar mass of each species by using their atomic masses
BA = 137.33g/mol
N = 14g/mol
O= 16g/mol
H = 1g/mol
S = 32g/mol
calculation
Ba(NO3)2 = Ba + 2N + 6O
= 137.33 + 2X 14 + 6 X 16
= 261.33g/mol
NH2SO3H = N + 3H + S+ 3O
=14 + 3X1 + 32 + 3X 16
= 97g/mol
Ba(NH2SO3)2 = Ba + 2N + 4H +2S +6O
= 137.33 + 2 X 14 + 4 X1 + 2X32 + 6 X 16
= 329.33g/mol
HNO3 = H + n + 3O
= 1 + 14 + 3 X 16
= 63g/mol
Answer:
3.94 L
Explanation:
From the question given above, the following data were obtained:
Mass of O₂ = 5.62 g
Volume of O₂ =?
Next, we shall determine the number of mole present in 5.62 g of O₂. This can be obtained as follow:
Mass of O₂ = 5.62 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mole of O₂ =?
Mole = mass / molar mass
Mole of O₂ = 5.62 / 32
Mole of O₂ = 0.176 mole
Finally, we shall determine the volume of 5.62 g (i.e 0.176 mole) of O₂ at STP. This can be obtained as follow:
1 mole of O₂ occupied 22.4 L at STP.
Therefore, 0.176 mole of O₂ will occupy = 0.176 × 22.4 = 3.94 L at STP.
Thus 5.62 g (i.e 0.176 mole) of O₂ occupied 3.94 L at STP
Answer: 500000 centimeters
Explanation: