I think it is Sodium,hypochlorite
The molar mass of the acid if 28. 5 ml of the koh solution is required to neutralize the sample is 90.23g/mol.
<h3>For the calculation of molarity of solution</h3>
Molarity = (moles of solute/volume of solution) × 1000
Given,
Molarity of KOH solution = 0.180 M
Volume of solution = 28.5 mL
0.180 = (moles of KOH/ 28.5) × 1000
Moles = (0.18× 28.5)/1000
= 0.00513 mol
<h3>Chemical equation for the reaction</h3>
HA + KOH ------- KA + H2O
1 moles of KOH reacts with 1 moles of HA.
So, 0.00513 moles of KOH react with 0.00513 moles of HA.
<h3>To calculate the molar mass for given number of moles</h3>
Number of moles= given mass/ Molar mass
Given,
Mass of HA = 0.462 g
Moles of HA = 0.00512 mol
0.00512 = 0.462/ Molar mass
Molar mass = 90.23 g/ mol.
Thus the molar mass of HA required to neutralize the 28.5 mL of KOH is 90.23g/mol.
learn more about molar mass or moles:
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Answer:
The answer to your question is 21.45 g of KBr
Explanation:
Chemical reaction
2K + Br₂ ⇒ 2KBr
14.4 ?
Process
1.- Calculate the molecular mass of bromine and potassium bromide
Bromine = 2 x 79.9 = 159.8g
Potassium bromide = 2(79.9 + 39.1) = 238 g
2.- Solve it using proportions
159.8 g of Bromine ------------ 238 g of potassium bromide
14.4 g of Bromine ------------ x
x = (14.4 x 238) / 159.8
x = 3427.2 / 159.8
x = 21.45g of KBr
start the balancing by writing down how many atoms there are per element. we’ll use this as an example:
C3H8 + O2 --> H2O + CO2
C = 3 C = 1
H = 8 H = 2
O = 2 O = 3
balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:
C3H8 + O2 --> H2O + (3)CO2
now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.
C = 3 C = 3
H = 8 H = 2
O = 2 O = 7
now let’s move on to the hydrogen atoms.
C3H8 + O2 --> H2O + 3CO2
you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.
C3H8 + O2 --> (4)H2O + 3CO2
now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.
C = 3 C = 3
H = 8 H = 8
O = 2 O = 10
now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10
C3H8 + (5)O2 --> 4H2O + 3CO2
the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.
C = 3 C = 3
H = 8 H = 8
O = 10 O = 10
