Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
Start by looking to the product there that doesn't have coefficient. Need to have exactly 2 Fe on the reactant side. So first coefficient is 2.. having a 2 there gives 2*3 = 6 H on the reactant side. Then in order to have 6 H on the product side the coefficient there needs to be 3. This gives 3 Ca on product side so there must be 3 Ca on reactant side.
coefficients:
2 + 3 --> 1 + 3
They have luster, good thermal and electrical conductivity, high densities and melting points, and are malleable. The only exception would be mercury most metals are solid at room temp
Set up a proportion os V1/n1 = V2/n2. V is the volume, n is the amount in MOLES, not grams. Convert the CO_2 to moles, then solve and find that the amount of N_2 should be the same amount of moles. Then use the molar mass of N_2 (28.02 grams/mole) to convert that amount of moles into grams. That's your answer.