Answer = c
Conservation of mass (mass is never lost or gained in chemical reactions), during chemical reaction no particles are created or destroyed, the atoms are rearranged from the reactants to the products.
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce 
g of (NH₄)₂S
25g of NH₃ will produce 
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S
The answer is the choice A
Answer:
9.69g
Explanation:
To obtain the desired result, first let us calculate the number of mole of N2 in 7.744L of the gas.
1mole of a gas occupies 22.4L at stp.
Therefore, Xmol of nitrogen gas(N2) will occupy 7.744L i.e
Xmol of N2 = 7.744/22.4 = 0.346 mole
Now let us convert 0.346 mole of N2 to gram in order to obtain the desired result. This is illustrated below:
Molar Mass of N2 = 2x14 = 28g/mol
Number of mole N2 = 0.346 mole
Mass of N2 =?
Mass = number of mole x molar Mass
Mass of N2 = 0.346 x 28
Mass of N2 = 9.69g
Therefore, 7.744L of N2 contains 9.69g of N2
The H3O+ in a 0.050M solution of Ba(OH)2 is calculated as below
write the equation for the dissociation of Ba(OH)2
Ba(OH)2 = Ba^2+ +2OH^-
calculate the OH- concentration
by use of mole ratio between Ba(OH)2 to OH^- which is 1:2 the concentration of OH = 0.050 x2 = 0.1 M
by use of the formula ( H3O+)(OH-) = 1 x10 ^-14
by making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-
substitute for OH-
H3O+ = (1 x10^-14 )/0.1
= 1 x10^-3 M
