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2 years ago

How much volume (in cm3) is gained by a person who gains 12.3 lb of pure fat?

Chemistry

1 answer:

maks197457 [2]2 years ago

4 0

Answer:

So we are given with the mass while we are asked for the amount of volume. So this means, we must need an information on the density. From literature, the density of human fat is 0.918 g/cm³. Convert grams to lb by the conversion that 1,000 g = 2.2 lbs.

Density = 0.918 g/cm³ * (2.2 lbs/1,000 g) = 0.0020196 lb/cm³

Volume = Mass/Density = 12.2 lb / 0.0020196 lb/cm³

Volume = 6,040.8 cm³

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A student measures the mass of an 8 cm^3 block of brown sugar to be 12.9 g. What is the density of

Oduvanchick [21]

Answer:

1.6125 g/cm^3

Explanation:

Density= mass/volume

4 0

2 years ago

1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate

Whitepunk [10]

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

$\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

$\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

6 0

2 years ago

This happens in diffusion. Choose the correct answer(s)

svlad2 [7]

The answer is d. particles move from an area of high concentration to low concentration.

8 0

2 years ago

Suppose that Daniel has a 3.00 3.00 L bottle that contains a mixture of O 2 O2 , N 2 N2 , and CO 2 CO2 under a total pressure of

Alenkinab [10]

Answer:

Partial pressure O₂ → 2.74 atm

Explanation:

Let's analyse the data given:

Volume → 3L

In the bottle there is a mixture of gases that contains, O₂, N₂ and CO₂.

Total pressure is 4.80 atm

Let's apply the Ideal Gases Law to determine the total moles of the mixture

P . V = n . R. T

4.80 atm . 3L = n . 0.082 . 273K

n = 4.80 atm . 3L / 0.082 . 273K → 0.643 moles

We apply the concept of mole fraction:

Mole fraction of a gas X = moles of gas X / Total moles

Mole fraction of a gas X = Partial pressure X / Total pressure

In a mixture, sum of mole fraction of each gas = 1

We determine mole fraction of N₂ → 0.230 / 0.643 = 0.357

We determine mole fraction of CO₂ → 0.350 atm / 4.80 atm = 0.0729

1 - mole fraction N₂ - mole fraction CO₂ = mole fraction O₂

1 - 0.357 - 0.0729 = 0.5701 → mole fraction O₂

We replace in the formula: Mole fraction O₂ = Partial pressure O₂ / 4.80 atm

0.5701 . 4.80 atm = Partial pressure O₂ → 2.74 atm

5 0

2 years ago

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A)The lower the greenhouse gases, the higher the ocean temperature.

HACTEHA [7]

The higher the greenhouse gases, the higher the ocean temperature

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6 0

2 years ago

Read 2 more answers