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3 years ago

If the observed value for a density is 0.80 g/mL and the accepted value is 0.70 g/mL what is the percent error?

Chemistry

1 answer:

kvv77 [185]3 years ago

4 0

Answer:

<h2>The answer is 14.29 %</h2>

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

actual density = 0.70 g/mL

error = 0.8 - 0.7 = 0.1

So we have

$P(\%) = \frac{0.1}{0.7} \times 100 \\ = 14.285714...$

We have the final answer as

Hope this helps you

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A reaction between substances Y and Z is

Anuta_ua [19.1K]

Answer : The value of activation energy for this reaction is 108.318 kJ/mol

Explanation :

The Arrhenius equation is written as:

$K=A\times e^{\frac{-Ea}{RT}}$

Taking logarithm on both the sides, we get:

$\ln k=-\frac{Ea}{RT}+\ln A$ ............(1)

where,

k = rate constant = $2.95\times 10^{-3}L/mol.s$

Ea = activation energy = ?

T = temperature = 435 K

R = gas constant = 8.314 J/K.mole

A = pre-exponential factor = $3.00\times 10^{+10}L/mol.s$

Now we have to calculate the value of rate constant by putting the given values in equation 1, we get:

$\ln (2.95\times 10^{-3}L/mol.s)=-\frac{Ea}{8.314J/K.mol\times 435K}+\ln (3.00\times 10^{10}L/mol.s)$

$Ea=108318.365J/mol=108.318kJ/mol$

Therefore, the value of activation energy for this reaction is 108.318 kJ/mol

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3 years ago

Which of the following properties decreases from left to right across a period? *

Anna71 [15]

Atomic radius atomic number ionization energy and then electronegativity

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3 years ago

) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H

Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

$pH = -log([H_{3}O^{+}])$

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

<u>Trimethyl ammonium</u>:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺

1.0 - x x x

$Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}$

$10^{-pKa} = \frac{x*x}{1.0 - x}$

$10^{-9.80}(1.0 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 0.097 = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01$

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

$Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$

$10^{-10} = \frac{x^{2}}{1.0 - x}$

$1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0$

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00$

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

$\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95$

Therefore, the difference in pH values is 4.95.

I hope it helps you!

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3 years ago

Crude oil is a mixture of many liquids. The crude oil enters the bottom of a tall column, where the mixture is heated. The subst

Yanka [14]

Boiling point ..................................................

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3 years ago

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PLEASE ANSWER: The density of a gas at STP is 0.75 g/L. What is the mass of this gas?

BartSMP [9]

Answer:

the mass of the glass is 2

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3 years ago