Answer:
1822.72 g
Explanation:
Applying,
n = R.M/M.M.................. Equation 1
Where n = number of moles of iron(II) nitrate, R.M = Reacting mass of iron(II) nitrate, M.M = molar mass of Iron(II) nitrate.
Make R.M the subject of the equation
R.M = n×M.M............. Equation 2
From the question,
Given: n = 6.33 mol
Constant: M.M of iron(II) nitrate = 287.95 g/mol
Substitute these values into equation 2
R.M = 6.33(287.95)
R.M = 1822.72 g
Hence the mass of iron(II) nitrate is 1822.72 g
Volume Ba(OH)2 = 23.4 mL in liters :
23.4 / 1000 => 0.0234 L
Molarity Ba(OH)2 = 0.65 M
Volume HNO3 = 42.5 mL in liters:
42.5 / 1000 => 0.0425 L
number of moles Ba(OH)2 :
n = M x V
n = 0.65 x 0.0234
n = 0.01521 moles of Ba(OH)2
Mole ratio :
<span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
</span>
1 mole Ba(OH)2 ---------------- 2 moles HNO3
0.01521 moles ----------------- moles HNO3
moles HNO3 = 0.01521 x 2 / 1
moles HNO3 = 0.03042 / 1
= 0.03042 moles HNO3
Therefore:
M ( HNO3 ) = n / volume ( HNO3 )
M ( HNO3 ) = 0.03042 / 0.0425
M ( HNO3 ) = 0.715 M
Answer:
Explanation:
Hello there!
In this case, when referring to single replacement reactions, it is crucial for us to figure out the formula of the starting reactants; thus, we know zinc is Zn and lead (IV) chloride is PbCl₄. In such a way, the reaction proceeds as follows:
Which must be balanced as shown below:
Regards!
A chemical reaction is a reaction that changes the molecular structure and is normally irreversible.
Mass percentage of sodium chloride(NaCl) in ocean waters = 3.5 %
That means 3.5 g sodium chloride(NaCl) is present for every 100 g of ocean water.
The given mass of sodium chloride(NaCl) is 45.8 g
Calculating the mass of ocean waters that would contain 45.8 g sodium chloride(NaCl):
= 1309 g ocean water
Therefore, 45.8 g sodium chloride is present in 1309 g ocean water.
