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inysia [295]
3 years ago
10

A student measures the mass of a sample as 9.67g. Calculate the percentage error, given that the correct mass is 9.82

Chemistry
2 answers:
max2010maxim [7]3 years ago
8 0

Answer:

1.5 %

Explanation:

Given data:

Theoretical value = 9.67 g

Actual value = 9.82 g

Percentage error = ?

Solution:

Formula

<em>percentage error = (Theoretical value - actual value / actual value) ×100</em>

percentage error = 9.67  - 9.82  /9.82  ×100

percentage error = -0.15/ 9.82 ×100

percentage error = -0.015×100

percentage error = 1.5 %

pishuonlain [190]3 years ago
5 0
<h3>Answer:</h3>

1.53%

<h3>Explanation:</h3>

We are given;

Experimental value as 9.67 g

Actual measurement as 9.82 g

We are required to calculate the percentage error;

  • Percentage error is given by dividing the absolute value of the difference between the experiment value and the actual value then dividing by the actual value and multiplying by 100.
  • That is;

% error = ((Experimental value -Actual value)÷ Actual value) × 100%

Difference between experimental value and actual value = 9.67 g - 9.82 g

                                                       = -0.15

Therefore, Absolute value = 0.15

Therefore;

% error = (0.15 ÷ 9.82 g) × 100

            = 1.527%

            = 1.53%

Thus, the percentage error is 1.53%    

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Hey there !

<span>Convert Joule to KJ :
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Kj = 53.69 * 0.001

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T = 49.09 / 0.05369

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You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph
zhuklara [117]

Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

  • C1V1 = C2V2

where

  • C1 is the concentration of stock solution
  • V1 is the volume of stock solution required to prepare a diluted solution
  • C2 is the concentration of the diluted solution prepared
  • V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

  • Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

6 0
2 years ago
When 481 J of energy is added to 18.3 g of aluminum at 17.4°C, the temperature increases
QveST [7]

Answer:

THE SPECIFIC HEAT OF ALUMINIUM IS 0.900 Jg^-1K^-1

Explanation:

To solve this question, follow the following steps listed below:

1. Write the formulae for heat change in a reaction

Heat = m c ΔT

2. Write out their values and convert to standard units.

m = mass of aluminum = 18.3g

Heat = 481 J

T2 = final temperature = 46.6°C = 273 + 46.6 = 319.6 K

T1 = initial temperature = 17.4 °C = 17.4 + 273 K = 290.4K

C = specific heat capacity of aluminum = ?

3. Solve for the specific heat of aluminium

c = Heat / m ΔT

c = 481 J / 18.3 g * ( 319.6 - 290.4) K

c = 481 / 18.3 * 29.2

c = 481 / 534.36

c = 0.900 Jg^-1K^-1

In conclusion, the specific heat capacity of aluminum is 0.900 Jg^-1K^-1.

8 0
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