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Verizon [17]
1 year ago
8

Suppose a 1.0 ml air bubble is trapped in the top of your buret and you do not notice it before measuring our liquid samples wit

h the buret. what percentage error would be introduced in a 35.0 ml sample of liquid if the air bubble comes out of the tip of the buret while the liquid sample is being transferred?
Chemistry
1 answer:
Nadusha1986 [10]1 year ago
7 0

The percentage error that would be introduced in a 35.0 mL sample of liquid is 2.86%

<h3>Data obatined from the question</h3>

From the question given, the following data were obtained:

  • Absolute error = 1 mL
  • True measurement = 35 mL
  • Perecentage error =?

<h3>How to determine the percentage error</h3>

The percentage error that would be introduced can be obtained as illustrated below:

Percentage error = (Absolute error / True measurement) × 100

Percentage error = (1 / 35) × 100

Percentage error = 2.86%

Thus, percentage error that would be introduced is 2.86%

Learn more about percentage error:

brainly.com/question/17207115

#SPJ1

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How much heat (kJ) is absorbed by 229.1 g of water in order for the temperature to increase from 25.00∘C to 32.50∘C?
hammer [34]

Answer:

(Q1) 9.42 kJ.

(Q2) 1.999 kJ

Explanation:

Heat: This is a form of Energy that brings about the sensation of warmth.

The S.I unit of Heat is Joules (J).

The heat of a body depend on the mass of the body, specific heat capacity, and temperature difference. as shown below

Q = cm(t₂-t₁) ........................ Equation 1

(Q1)

Q = cm(t₂-t₁)

Where Q = amount of heat absorbed, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 229.1 g = 0.2991 kg, t₁ = 25.0 °C, 32.50 °C

Constant: c = 4200 J/kg.°C

Substituting into equation 1

Q = 0.2991×4200(32.5-25)

Q = 1256.22(7.5)

Q = 9421.65 J

Q = 9.42 kJ.

Hence the heat absorbed = 9.42 kJ

(Q2)

Q = cm(t₂-t₁)

Where Q = amount of heat required, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature.

Given: m = 34 g = 0.034 kg, t₁ = 9 °C, t₂ = 23 °C

Constant: c = 4200 J/kg.°C

Q = 0.034×4200(23-9)

Q = 142.8(14)

Q = 1999.2 J

Q = 1.999 kJ.

Thus the Heat required = 1.999 kJ

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Suppose that there is an air sample and a water sample, each of the same mass and each with an initial temperature, Tinitial, of
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