Explanation:
It is given that volume of is 300 mL and molarity is 0.200 M.
Volume of NaCl is 200 mL and molarity is 0.050 M.
The chemical reaction will be as follows.
for is given as .
As, molarity is number of moles present in liter of solution.
Hence, moles of (aq) will be calculated as follows.
moles of (aq) =
= 0.06 mol
=
= 0.120 M
Mole of =
= 0.010 M
Now, Q =
=
=
As, Q < hence, there will be no formation of precipitate.
55.36 is the mass of Calcium Bromide is needed to make 0.500 L of 0.554 M solution.
Explanation:
Data given:
mass of Ca = ?
molarity of the calcium bromide solution = 0.554 M
Volume of the calcium bromide solution= 0.5 L
First the moles of calcium bromide is calculated as:
molarity =
number of moles = 0.554 x 0.5
number of moles = 0.277 moles of calcium bromide
atomic mass of calcium bromide = 199.89 grams/mole
mass = atomic mass x number of moles
mass of calcium bromide = 199.89 x 0.277
= 55.36 grams
mass of calcium bromide required is 55.36 grams to make solution of 0.554 M in a solution of 0.5 litres.
The organism would no longer grow.
The reaction is at equilibrium at 1,000 K. The equilibrium constant of the reaction is 3.90. At equilibrium, the concentrations are as follows.
[CO] = 0.30 M
[H2] = 0.10 M
[H2O] = 0.020 M
What is the equilibrium concentration of CH4 expressed in scientific notation?
answer is b
Answer:
Therefore 9 moles of nitrogen are required to produce 18 moles of ammonia.