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marishachu [46]
3 years ago
12

A zinc slug comes from a science supply company with a stated mass of 5.000 g. A student weighs the slugthree times, collecting

the following values: 4.891 g, 4.901 g, and 4.890 g. Are the student’s values accurate?Are they precise (consider both meanings)?
Chemistry
1 answer:
soldi70 [24.7K]3 years ago
8 0

Answer: The student’s values are accurate as well as precise.

Explanation:

Precision refers to the closeness of two or more measurements to each other.

For Example: If you weigh a given substance three times and you get same value each time. Then the measurement is very precise.

Accuracy refers to 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 49 kg and another person weighed 48 kg. Then, the weight measured by first person is more accurate.

Given: Mass = 5.000 g

Mass weighed by A has values 4.891 g , 4.901 g and 4.890. Thus the average value is \frac{4.891+4.901+4.890}{3}=4.894

Thus as the measured value is close to the true value, the student’s values are accurate and as the values are close to each other, the measurement is precise.

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What is the total mass of Hydrogen in each of the molecules?
frutty [35]

Explanation:

a) In 1 mole of methane there are 4 moles of hydrogen atom

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of methane = 4 × 1 g = 4 g

b) In 1 mole of chloroform there are 1 mole of hydrogen atom

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of methane = 1× 1 g = 1 g

c) In 1 mole of C_{12}H_{10}O_{16} there are 10 moles of hydrogen atom

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of C_{12}H_{10}O_{16} = 10 × 1 g = 10 g

d)In 1 mole of CH_3CH_2CH_2CH_2CH_3 there are 12 moles of hydrogen atom.

CH_3CH_2CH_2CH_2CH_3=C_5H_{12}

Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of CH_3CH_2CH_2CH_2CH_3 = 12 × 1 g = 12 g

3 0
3 years ago
In a reaction vessel, 17.6 g of solid chromium(III) oxide, Cr2O3, was allowed to react with excess carbon tetrachloride in the f
notsponge [240]

Answer:

72.53% is the yield of CrCl3

Explanation:

Given

Reaction:

Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)

CCl4 is in excess and 17.6g  Cr2O3 present

The reaction yields 26.6g of CrCl3

To Find:

% yields of the reaction

Also given

Molar mass of CrCl3 = 158.35g/mol

Molar mass of Cr2O3 = 152.00 g/mol

By the stoichiometry of the reaction

1 mole of Cr2O3 gives  2 moles of CrCl3

0r

1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3

= 1 52 g of Cr2O3 gives 316.70 g of CrCl3

    17.6 g of Cr2O3 gives  (17.6÷152) × 316.70 g CrCl3

= 36.67 g CrCl3

but actual yield is only 26.6g

so % yield is (26.6 ÷÷ 36.67) × 100

= 72.53% is the yield of CrCl3

8 0
3 years ago
What is an ideal gas? how many ideal gases are there in the universe?
tiny-mole [99]
Ideal gases are hypothetical gases whose molecules occupy negligible space and have no interactions, and that consequently obeys the gas laws exactly.
Not exactly sure about the amount...
I hope this helps! :)
6 0
3 years ago
A scientist investigates a new substance and finds that at 35° C, 146.2 grams will dissolve in 200.0 grams of water. Which simil
Nimfa-mama [501]
The scientist's results is that at a temperature of 35<span>°C, the solubility of the substance in water is 146.2 grams in 200 grams of water. There isn't really a different method to determine the solubility of a substance in water. Another procedure could be that a lesser amount of the substance is used and the water required to dissolve it is determined. The solubility of the substance based on the two procedures can then be compared.</span>
5 0
2 years ago
Using the van der waals equation, the pressure in a 22.4 L vessel containing 1.50 mol of chlorine gas at 0.00 c is____________at
MatroZZZ [7]

Answer:

D. 1.48atm

Explanation:

Van der waals equation is given as:

(P +an²/v²) (v - nb) = nRT

Where;

P = pressure (atm)

V = volume (L)

R = gas constant (0.0821 Latm/molK)

a and b = gas constant specific to each gas

T = temperature (K)

n = number of moles

According to the given information; V = 22.4L, T = 0.00°C (273.15K), R = 0.0821 Latm/molK, a = 6.49L^2-atm/mol^2, b = 0.0562 L/mol, n = 1.5mol

Hence;

(P + 6.49 × 1.5²/22.4²) (22.4 - 1.5×0.0562) = 1.5 × 0.0821 × 273.15

(P + 6.49 × 2.25/501.76) (22.4 - 0.0843) = 33.638

(P + 0.0291) (22.316) = 33.638

22.316P + 0.649 = 33.638

22.316P = 33.638 - 0.649

22.316P = 32.989

P = 32.989/22.316

P = 1.478

P = 1.48atm

6 0
3 years ago
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