Yes,sr8 facts, mhm, agree
The mixture contains:
CaCO3 + (NH4)2CO3 in which the amount of carbonate CO3 = 60.7% by mass
Let, the total mass = 100 grams
Mass of CaCO3 = x grams
Mass of (NH4)2CO3 = y grams
Thus, x + y = 100 ------------(1)
Mass of CO3 = 60.7% = 60.7 g
Molar mass of CO3 = 60 g/mol
Total # moles of CO3 = 60.7 g/60 g.mol-1 = 1.012 moles
The total moles of CO3 comes from CaCO3 and (NH4)2CO3. Therefore,
moles CaCO3 + moles (NH4)2CO3 = 1.012
mass CaCO3/molar mass CaCO3 + mass (NH4)2 CO3/molar mass = 1.012
x/100 + y/96 = 1.012---------(2)
based on equation 1 we can write: y = 100-x
x/100 + (100-x)/96 = 1.012
x = 71.2 g
Mass of CaCO3 = 71.2 g
Answer:
She should have first notified the teacher.
The equation that we will use to solve this question is:
M1*V1 = M2*V2
where:
M1 is the initial concentration = 9.13*10^-2 M
V1 is the initial volume = 80 ml
M2 is the final concentration = 1.6*10^-2 M
V2 in the final volume that we need to find
Substitute with the values given in the equation to calculate the final volume as follows:
(9.13*10^-2*80) = (1.6*10^-2) * V2
V2 = 456.5 ml
6 protons
4 electrons in 1st shell
2 valence electrons 2nd shell
