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:
A crack in earth's crust is called a fault also known as a "fracture"
To determine the limiting reactant, we first calculate the number of moles of the reactant.
7.00 g H2 ( 1 mol / 2.02 g ) = 3.47 mol H2
70.00 g N2 ( 1 mol / 28.02 g) = 2.50 mol N2
From the balanced reaction, we see that there is a 3 is to 1 ratio of the reactants. So, for every 3 moles of H2 we need 1 mole of N2. Given the amount of reactant, for 2.50 moles of N2 we need 7.5 moles of H2 which obviously we have a lesser amount. Hydrogen is the limiting reactant.
Answer:
43.75 ml
Explanation:
Given that the equation of the reaction is;
2HNO3(aq) + Ca(OH)2(aq) ---> Ca(NO3)2(aq) + 2 H20(l)
Concentration of acid CA= 0.05 M
Concentration of base CB = 0.02 M
Volume of acid VA = 35.00ml
Volume of base VB= ???
Number of moles of acid NA= 2
Number of moles of base NB=1
From
CAVA/CBVB= NA/NB
Making VB the subject of the formula;
VB= CAVANB/CBNA
VB= 0.05 × 35 × 1/ 0.02 × 2
VB=1.75 /0.04
VB= 43.75 ml
