Explanation:
2nacl+baso4
the orginal equation is
Na2so4+Bacl2
then na react with cl and ba react with so4
Molar mass is the ratio of the mass to that amount of the substance. The mass of the barium nitrate in the formula unit is 23.0 grams.
<h3>What is mass?</h3>
The mass of a substance is the product of the molar mass of the compound and the number of moles of the compound.
Given,
Molar mass of barium nitrate = 261.35 g/mol
If,
have a mass of 261.35 g/mol then,
formula units will have a mass of,

Therefore, option C. 23.0 gm is the mass of barium nitrate.
Learn more about mass here:
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Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
The abundance of first isotope is 69.15 %
The abundance of second isotope is 30.85 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope:
% = x %
Mass = 62.9296 u
For second isotope:
% = 100 - x
Mass = 64.9278 u
Given, Average Mass = 63.546 u
Thus,
Solving for x, we get that:
x = 69.15 %
<u>The abundance of first isotope is 69.15 %</u>
<u>The abundance of second isotope is 100 - 69.15 % = 30.85 %</u>