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Feliz [49]
3 years ago
6

A student is asked to determine the molarity of a strong base by titrating it with 0.250 M solution of H2SO4. The students is in

structed to pipet a 20.0 mL portion of the strong base solution into a conical flask, to add two drops of an indicator that changes at pH = 7, and to dispense the standard H2SO4 solution from a buret until the solution undergoes a permanent color change. The initial buret reading is 5.00 mL and the final reading is 30.00 mL at the equivalence point.Which procedural error will result in a strong base molarity that is too high?Using an indicator that changes at pH = 5 rather than at pH = 7Using a buret with a tip filled with air rather than the H2SO4 solutionUsing 4 drops of indicator rather than the recommended 2 dropsUsing a conical flask that contains several drops of H2O
Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

i working on this too

i might be able to tell u the right answer when im done

Explanation:

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Arlecino [84]

Answer:

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Explanation:

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Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat
Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

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Hi can someone hellp this is grade 7
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Explanation:

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