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timofeeve [1]
3 years ago
10

10.00 mL of 0.10 M NaOH is added to 25.00 mL of 0.10 M HCl during a titration

Chemistry
1 answer:
bekas [8.4K]3 years ago
3 0

Answer:

Explanation:

10 mL = .01 L .

25 mL = .025 mL .

10 mL of .1 M NaOH will contain .01 x .1 = .001 moles

25 mL of .1M HCl will contain .025 x .1 = .0025 moles

acid will neutralise and after neutralisation moles of acid remaining

= .0025 - .001 = .0015 moles .

Total volume = .01 + .025 = .035 L

concentration of remaining HCl = .0015 / .035

Option D is correct.

= .042857 M

= 42.857 x 10⁻³ M .

pH = - log [42.857 x 10⁻³]

= 3 - log 42.857

= 3 - 1.632

= 1.368 .

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A chemical equation does not give information about the following:

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8 0
3 years ago
During a hurricane, what effect can the ocean have on the beach?
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C is the answer I’m pretty sure
4 0
3 years ago
A steel cylinder of oxygen is being store in a room at 25.0 °C under a pressure of 1250 atm. What pressure would be exerted by t
Rudik [331]

Answer:

P₂ = 1312.88 atm

Explanation:

Given data:

Initial temperature = 25°C

Initial pressure = 1250 atm

Final temperature = 40°C

Final pressure = ?

Solution:

Initial temperature = 25°C (25+273.15 = 298.15 K)

Final temperature = 40°C ( 40+273.15 = 313.15 k)

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1250 atm / 298.15 K = P₂/313.15 K

P₂ = 1250 atm × 313.15 K / 298.15 K

P₂ = 391437.5 atm. K /298.15 K

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6 0
3 years ago
A researcher studying the nutritional value of a new candy places a 4.90 g sample of the candy inside a bomb calorimeter and com
snow_lady [41]

Answer:

449730.879 cal/g

Explanation:

Given data:

Mass of sample = 4.9 g

Change in temperature  = 2.08 °C  (275.23 k)

Heat capacity of calorimeter = 33.50 KJ . K⁻¹

Solution:

C(candy) = Q/m

Q = C (calorimeter) × ΔT

C(candy) = C (calorimeter) × ΔT / m

C(candy) =  33.50 KJ . K⁻¹ × 275.23 K / 4.90 g

C(candy) = 9220.205 KJ / 4.90 g

C(candy) =  1881.674 KJ / g

It is known that,

1 KJ /g = 239.006 cal/g

1881.674 × 239.006 = 449730.879 cal/g

8 0
3 years ago
When 5.00 mol of benzene is vaporized at a constant pressure of 1.00 atm and at its normal boiling point of 80.1°C, 169.5 kJ are
NNADVOKAT [17]

Answer:

∆E=155kJ

Explanation:

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3 0
3 years ago
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