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Grace [21]
3 years ago
8

Jamie is measuring the mass of one mole of carbon. She measures the mass 4 times, achieving the following measurements: 11.86 g,

11.87 g, 11.89 g, and 11.90 g. If the accepted value of one mole of carbon is 12.11 g, how can Jamie's measurements best be described
Chemistry
1 answer:
azamat3 years ago
7 0

This is an incomplete question, here is a complete question.

Distinguish between Accuracy and Precision Question Jamie is measuring the mass of one mole of carbon. She measures the mass 4 times, achieving the following measurements: 11.86 g, 11.87 g, 11.89 g, and 11.90 g. If the accepted value of one mole of carbon is 12.11 g, how can Jamie's measurements best be described?

Select the correct answer below:

Jamie's measurements are both precise and accurate,

Jamie's measurements are accurate but not precise.

Jamie's measurements are precise but not accurate,

Jamie's measurements are neither precise nor accurate,

Answer : The correct option is, Jamie's measurements are precise but not accurate.

Explanation :

Accuracy : It is defined as the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 50 kg and one person weighed 48 kg and another person weighed 55 kg. Then, the weight measured by first person is more accurate.

Precision : It is defined as the closeness of two or more measurements to each other.

For Example: If you weigh a given substance five times and you get 1.8 kg each time. Then the measurement is said to be precise.

Level of precision is determined by the maximum number of decimal places.

As we are given that:

The measurements are, 11.86, 11.87, 11.89 and 11.90

The accepted value is, 12.11

The average of these values =  \frac{11.86+11.87+11.89+11.90}{4}=11.88

From this we conclude that, these measurements are close to each other that means they are precise. But the accepted value are not equal to average of four values that means they are not accurate.

Hence, the correct option is, Jamie's measurements are precise but not accurate.

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Find the percentage composition of a compound that contains 1.94 g of carbon, 0.48 g of hydrogen, and 2.58 g of sulfur.
Svetradugi [14.3K]

Answer : The percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

Explanation :

To calculate the percentage composition of element in sample, we use the equation:

\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100

Given:

Mass of carbon = 1.94 g

Mass of hydrogen = 0.48 g

Mass of sulfur = 2.58 g

First we have to calculate the mass of sample.

Mass of sample = Mass of carbon + Mass of hydrogen + Mass of sulfur

Mass of sample = 1.94 + 0.48 + 2.58 = 5.0 g

Now we have to calculate the percentage composition of a compound.

\%\text{ composition of carbon}=\frac{1.94g}{5.0g}\times 100=38.8\%

\%\text{ composition of hydrogen}=\frac{0.48g}{5.0g}\times 100=9.6\%

\%\text{ composition of sulfur}=\frac{2.58g}{5.0g}\times 100=51.6\%

Hence, the percentage composition of carbon, hydrogen and sulfur in a compound is, 38.8 %, 9.6 % and 51.6 % respectively.

3 0
3 years ago
How many moles of zinc are there in 0.890 g of zinc?
NNADVOKAT [17]

Answer:

There are  

4.517

⋅

10

23

atoms of Zn in 0.750 mols of Zn.

Explanation:

Since we know that there are  

6.022

⋅

10

23

atoms in every mole of a substance (Avogadro's Number), there are  

6.022

E

23

⋅

0.750

atoms of Zn in 0.750 mols of Zn.

4 0
2 years ago
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How many grams of oxygen are in 5.3 moles of lactose? (Using percent composition)
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2 years ago
A sample of a compound is decomposed in the laboratory and produces 330 g carbon, 69.5 g hydrogen, and 220.2 g oxygen. Calculate
Zarrin [17]

Answer: Empirical formula is C_2H_5O

Explanation: We are given the masses of elements present in a sample of compound. To evaluate empirical formula, we will be following some steps.

<u>Step 1 :</u> Converting each of the given masses into their moles by dividing them by Molar masses.

Moles=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of Carbon = 12.0 g/mol

Molar mass of Hydrogen = 1.0 g/mol

Molar mass of Oxygen = 16.0 g/mol

Moles of Carbon = \frac{330g}{12g/mol}=27.5moles

Moles of Hydrogen = \frac{69.5g}{1g/mol}=69.5moles

Moles of Oxygen = \frac{220.2g}{16g/mol}=13.76moles

<u>Step 2: </u>Dividing each mole value by the smallest number of moles calculated above and rounding it off to the nearest whole number value

Smallest number of moles = 13.76 moles

\text{Mole ratio of Carbon}=\frac{27.5moles}{13.76moles}=1.99\approx 2

\text{Mole ratio of Hydrogen}=\frac{69.5moles}{13.76moles}=5.05\approx 5

\text{Mole ratio of Oxygen}=\frac{13.76moles}{13.76moles}=1

<u>Step 3:</u> Now, the moles ratio of the elements are represented by the subscripts in the empirical formula

Empirical formula becomes = C_2H_5O

7 0
3 years ago
Enter the chemical formula of a binary molecular compound of hydrogen and a Group 6A element that can reasonably be expected to
NARA [144]

Answer:

Any binary molecular compound of hydrogen and a Group 6A element above Selenium will be less acidic, so water and dihydrogen sulfide are less acidic in aqueous solution than hydrogen selenide.

Explanation:

Going down in a group increases the atomic radius and a greater atomic radius implyes greater ionic radius.  

When ionization takes place in these compounds they yelds protons (hidrogen ion) and an lewis base (anion). The greater the ionic radius the greater its stability, thus the periodic tendency is increaing the acidity of binary hidrogen compounds when going down a group. On the other hand going up a group decreases acidity, so any molecular compound of hydrogen and a Group 6A element above Selenium will be less acidic, so water and dihydrogen sulfide are less acidic in aqueous solution than hydrogen selenide.

3 0
3 years ago
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