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3 years ago

A scientist measures a substance to be 0.6 grams. Calculate the percent of error in the measurement.

1 answer:

7 0

Percent error is an expression that is used as a tool for determining the precision of your calculations. It is calculated by the absolute value of the difference of experimental and theoretical values over the theoretical values multiplied by 100%. Hope this helps.

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When forming an ion, this sodium atom will lose 1 electron, as it is the easiest way to form a full shell, as opposed to gaining 7 electrons.

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3 years ago

Weathering and Erosion Unit Test

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2 years ago

The first step in deforestation in the Amazon is _______. A. Logging b. Farming c. Ranching d. Road building.

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Explanation: The answer is A. logging

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3 years ago

Kia is doing an experiment in science lab. She is given a beaker containing 100 g of liquid. The beaker has markings on the side

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3 years ago

Read 2 more answers

A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a

Semenov [28]

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

We are given:

Mass percentage of $CH_4$ = 20 %

So, mole fraction of $CH_4$ = 0.2

Mass percentage of $C_2H_4$ = 30 %

So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

where,

$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$

Hence, the correct answer is Option d.

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4 years ago