Answer:
-4
Explanation:
In 0.0034 the zeros are preceding numberd and after the decimal point.
We know that molarity = mol/L, so:
1.5 M = x mol/0.345 L
1.5 M * 0.345 L = x mol => 0.5175 mol
0.5175 mol/0.250 L = 2.07 M
Your new molarity of the solution will be 2.07 M.
5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.
Explanation:
The balanced equation:
2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)
data given:
volume of AgNO3 = 0.2 L
molarity of AgNO3 = 0.200 M
atomic weight of AgCl= 143.32 gram/mole
from the formula, number of moles can be calculated
Molarity = 
number of moles of AgNO3 = 0.04
From the reaction:
2 moles of AgNO3 reacts to form 2 moles of AgCl
0.04 moles of AgNO3 reacts to form x mole of AgCl
= 
= 0.04 moles of AgCl is formed
mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl
mass of AgCl = 0.04 x 143.32
= 5.732 grams of AgCl is formed.
Answer:
d
Explanation:
the answer because decomposition means breaking down or taking apart