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iren [92.7K]
2 years ago
12

What is the molarity of a solution of h3po4 if 10.2 ml is neutralized by 53.5 ml of 0.20m koh what is the molarity of a solution

of h3po4 if 10.2 ml is neutralized by 53.5 ml of 0.20m koh 0.67m h3po4 0.15m h3po4 0.35m h3po4 0.37m h3po4
Chemistry
1 answer:
Anna35 [415]2 years ago
8 0

The molarity of the solution of H₃PO₄ needed to neutralize the KOH solution is 0.35 M

<h3>Balanced equation </h3>

H₃PO₄ + 3KOH —> K₃PO₄ + 3H₂O

From the balanced equation above,

  • The mole ratio of the acid, H₃PO₄ (nA) = 1
  • The mole ratio of the base, KOH (nB) = 3

<h3>How to determine the molarity of H₃PO₄ </h3>
  • Volume of acid, H₃PO₄ (Va) = 10.2 mL
  • Molarity of base, Ca(OH)₂ (Mb) = 0.2 M
  • Volume of base, Ca(OH)₂ (Vb) = 53.5 mL
  • Molarity of acid, H₃PO₄ (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 10.2) / (0.2 × 53.5) = 1 / 3

(Ma × 10.2) / 10.7 = 1 / 3

Cross multiply

Ma × 10.2 × 3 = 10.7

Ma × 30.6 = 10.7

Divide both side by 30.6

Ma = 10.7 / 30.6

Ma = 0.35 M

Learn more about titration:

brainly.com/question/14356286

#SPJ1

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In a synthesis reaction, one reactant contains 346 J of chemical energy, and one reactant contains 153 J of chemical energy. The
Eduardwww [97]

Answer:

64J of energy must have been released.

Explanation:

Step 1: Data given

One reactant contains 346 J of chemical energy, the other reactant contains 153 J of chemical energy.

The product contains 435 J of chemical energy.

Step 2:

Since the energy is conserved

Sum of energy of Reactants = Energy of Products

Sum of energy of Reactants = 346 J + 153 J = 499 J

The energy of the product = 435 J

435 < 499

This means energy must have been lost as heat.

Step 3: Calculate heat released

499 J - 435 J = 64 J

64J of energy must have been released.

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3 years ago
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solniwko [45]
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During the drying cycle, clothes soaked in liquid PERC at 300 K at 1 atm are tumbled in a stream of warm air at 330 K at 1 atm t
djyliett [7]
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6 0
3 years ago
How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

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7 0
2 years ago
Which of the Atoms shown has an atomic number four
dimulka [17.4K]

Answer:

B

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Atomic # = Protons

it says 4 p in the inside of the orbital

4 0
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