The relative molecular mass of acid A : 50 g/mol
<h3>Further explanation</h3>
Given
40.0 cm³(40 ml) of 0.2M sodium hydroxide
0.2g of a dibasic acid
Required
the relative molecular mass of acid A
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence(number of H⁺/OH⁻)
NaOH ⇒ n = 1
Dibasic acid = diprotic acid (such as H₂SO₄)⇒ n = 2
mol = M x V
Input the value in the formula :(1 = NaOH, 2=dibasic acid)
0.2 x 40 x 1 = M₂ x V₂ x 2
M₂ x V₂ = 4 mlmol = 4.10⁻³ mol ⇒ mol of Acid A
The relative molecular mass of acid A (M) :
Start with the 19.7 mol HNO3. use dimensional analysis to correctly convert from mol HNO3 to gram H2O. so, it should look similar to 19.7 mol HNO3 x (2 mol H2O/6 mol HNO3) x (18 g H2O/1 mol H2O)
the first parenthesis’ numbers were received from the balanced equation (for every 6 mol HNO3, 2 mol H2O formed). the second is converting from moles to grams by using the molar mass of H2O (1+1+16). you should get 709.2/6. once you divide those, the answer should be 118.2 g H2O. I’m not sure if your computer requires you to use the exact answer or stop at the correct number of significant digits, but if it does then it might just be 118. g H2O.
I need more info this is not enough but I think it’s 2 of electron and 7 nitrogen which is acetic mass
The chemical formula for carbon and chlorine is CCl4 (carbon tetrachloride).
Answer : The density of air in 
and 
is, 
and 
respectively.
Explanation :
where,
P = pressure of air = 1 atm
V = volume of air
T = temperature of air = 297 K
The conversion used for the temperature from Fahrenheit to degree Celsius is:
The conversion used for the temperature from degree Celsius to Kelvin is:
n = number of moles
m = mass of air
M = average molar mass of air = 28.97 g/mole
= density of air = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the above formula, we get:
Now we have to calculate density in .
Conversion used :
So,
The density of air in is,
Now we have to calculate density in .
Conversion used :
So,
The density of air in is,
