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2 years ago

How much volume (in cm^3)is gained by a person who gains 11.8 lb of our fat? Human fat has a density of 0.918 g/cm^3

Chemistry

1 answer:

Zinaida [17]2 years ago

3 0

The volume (in cm³) gained by a person who gains 11.8 lb of fat is 5830.49 cm³

<h3>What is density? </h3>

The density of a substance is simply defined as the mass of the subtance per unit volume of the substance. Mathematically, it can be expressed as

Density = mass / volume

<h3>How to convert pounds to grams </h3>

1 lb = 453.592 g

Therefore,

11.8 lb = 11.8 × 453.592

11.8 lb = 5352.3856 g

<h3>How to determine the volume </h3>

Mass = 5352.3856 g
Density = 0.918 g/cm³
Volume =?

Volume = mass / density

Volume = 5352.3856 / 0.918

Volume = 5830.49 cm³

Learn more about density:

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8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?

Lynna [10]

Answer:

the frequency of photons $v = 1.509\times10^{39}Hz$

Explanation:

Given: first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

$\Delta E_1 = 1000KJ/mol$

We know that plank's constant

$h = 6.626\times10^{-34} Js$

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

$= \frac{10^6J}{6.626\times10^{-34} Js}$

$v = 1.509\times10^{39}Hz$

Hence, the frequency of photons $v = 1.509\times10^{39}Hz$

6 0

3 years ago

If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),

garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0

3 years ago

In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro

Scorpion4ik [409]

Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}$

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

Molar mass of NaBr = 103 g/mole

Mass of water = 315 g

Now put all the given values in the above formula, we get:

$0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}$

$\text{Mass of NaBr}=18.3g$

Thus, the mass of sodium bromide added should be, 18.3 grams.

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3 years ago

Heat is being removed from a substance, but its temperature does not change. What is

ololo11 [35]

So its temperature will not rise, since kinetic energy of molecules remains the same. The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature (e g. From solid to liquid at melting point or from liquid to gas at boiling point) is termed as its latent heat.

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2 years ago

Investigate Weathering and Erosion 1. Describe at least one part of the experimental procedure that you thought was essential to

zvonat [6]

Answer:

Controlling the environment is the most key procedures for getting good results.

Explanation:

The control environment for an experiment is the essential part for getting good results. In control environment, there is no or less chances of disruption

from the external environment which can cause the results of the data more acceptable. So the scientists prefers laboratory for performing experiment as compared to outer environment. So in my opinion for getting better results, the control environment is the most necessary experimental procedure.

4 0

3 years ago

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