Initial Volume (VI)Final Volume (Vf) =Object's volume =
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Answer : The mass of nitric acid is, 214.234 grams.
Solution : Given,
Moles of nitric acid = 3.4 moles
Molar mass of nitric acid = 63.01 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of nitric acid.
Therefore, the mass of nitric acid is, 214.234 grams.
Answer:
Explanation:
The total mole of added = 15 × 1 × 10⁻³
= 15 × 10⁻³ mole
= 15 mmoles
the number of moles of NaOH added in order to neutralize the excess acid = 0.5905 × 29.393 = 17.36 mmoles
the equation for the reaction is:
2NaOH + -------->
i.e
2 moles of NaOH react with
1 moles of NaOH will react with 1/2
17.36 mmoles of NaOH = 1/2 × 17.36 mmoles of
= 8.68 mmoles
Number of moles of that react with MCO₃ = Total moles of
added - moles of
reacted with NaOH
= (15 - 8.68) mmoles
= 6.32 mmoles
+
------>
+
+
1 mole of react with 1 mole of
6.32 mmoles of = 6.32 mmoles of
number of moles of = 6.32 10⁻³ moles
mass of (carbonate salt) = 0.533 g
molar mass of = (M+60)g/mol
We all know that ;
number of moles = mass/molar mass
Then:
6.32 10⁻³ = 0.533 / (M+60)
(M+60) = 0.533/ 6.32 10⁻³
M + 60 = 84.34
M = 84.34 - 60
M = 24.34
Thus the element with the atomic mass of 24.34 is Chromium
The chemical formula for the compound is :