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3 years ago

How to solve 20.0 mL/ 5.7 g with significant figures and including units

Chemistry

1 answer:

Nutka1998 [239]3 years ago

7 0

$Question$

How to solve 20.0 mL/ 5.7 g with significant figures and including units?

Answer:

$p = 0.25 g/cm3$

Hope this Helps!!

$xXxAnimexXx$

The formula for density is the mass of an object divided by its volume. In equation form, that's d = m/v , where d is the density, m is the mass and v is the volume of the object. The standard units are kg/m³.

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At 20 degrees Celsius the density of mercury is 13.6 g/cm cubed. What is the mass of 50.0 ml of mercury at 20 degrees Celsius?

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680g of mercury is the mass of the sample of mercury.

Explanation:

Density is an unit used to relates the mass of an object per unit of volume.

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50.0mL * (13.6g / mL) =

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Answer:

See explanation

Explanation:

Lanark proposed a theory of evolution hinged on use and disuse of body parts. This is a theory that upholds the transfer of acquired traits to offsprings. Darwin's theory of evolution was rather based on natural selection or 'survival of the fittest'. Darwin's theory has more evidence backing it compared to Lamarck's theory. The field of genetics which appeared long after Darwin's death confirmed that inherited traits are indeed passed from parents to offsprings. These inherited traits are usually those that give organisms an edge over others in their environment.

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From the activity, "Hiding in the Background", individual worms spent their entire life cycle in either green or beige colour. Their colour didn't change as a result of the environment where they lived, rather the green earthworms were more in number, not particularly because of any specific changes the earthworms themselves had undergone, but because most of their predators concentrated on eating the beige worms. As a result of this, there were many green worms when the next generation began compared to the number of beige worms. Hence, there were more green earthworms.

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In a laboratory, 1.55mg of an organic compound containing carbon, hydrogen, and oxygen is burned for analysis. This combustion r

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Step-by-step explanation:

1. Calculate the mass of each element

Mass of C = 1.45 mg CO₂ × (12.01 mg C/44.01 mg CO₂) = 0.3957 mg C

Mass of H = 0.89 mg H₂O × (2.016 mg H/18.02 mg H₂O) = 0.0996 mg H

Mass of O = Mass of compound - Mass of C - Mass of H = (1.55 – 0.3957 – 0.0996) mg = 1.055 mg

2. Calculate the moles of each element

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Moles of H = 0.0996 mg H × 1 mmol H/1.008 mg H = 0.0988 mol H

Moles of O = 1.055 mg O × 1 mmol O/ 16.00 mg O = 0.06592 mol O

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

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H: 0.0988/0.032 95 = 2.998

O: 0.065 92/0.032 95 = 2.001

4. Round the ratios to the nearest integer

C:H:O = 1:3:2

5. Write the empirical formula

The empirical formula is CH₃O.

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