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3 years ago

A 75 mL solid is estimated by a student to be 81 mL. Calculate the percent error.

Chemistry

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

error = 81 - 75 = 6 mL

actual volume = 75 ml

We have

$p(\%) = \frac{6}{75} \times 100 \\ = \frac{2}{25} \times 100 \\ = 2 \times 4 \\ = 8 \: \: \: \: \: \: \: \:$

We have the final answer as

Hope this helps you

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4 years ago

Identifying Characteristics of Atoms

Anon25 [30]

Explanation:

The number of protons in an atom is the atomic number.

Mass number is the number of protons plus neutrons in an atom.

Mass number = number of protons + number of neutrons

Atomic number = number of protons

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Number Number Atomic Mass Symbol

Protons neutrons Number Number

A 7 B 15 C

D E 26 56 F

A, number of protons = B

Mass number = protons + neutrons

15 = protons + 7

Protons = 15 - 7 = 8

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3 years ago

Write the expression for the equilibrium constant Kp for the following reaction. (Enclose pressures in parentheses and do NOT wr

Liono4ka [1.6K]

Answer and Explanation:

For the following balanced reaction:

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

$Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}$

Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).

6 0

3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)