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Serhud [2]
3 years ago
15

A serving of soup contains 5.0 g of fat, 20.0 g of carbohydrates, and 3.0 g of protein. Determine how many kilocalories (kcal) a

re contained in this serving.
Chemistry
1 answer:
marta [7]3 years ago
4 0

<u>Answer:</u> The amount of kilocalories contained in the given serving of soup is 0.137 kCal

<u>Explanation:</u>

<u>As per the USDA:</u>

Carbohydrates provide 4 calories per gram, protein provides 4 calories per gram, and fat provides 9 calories per gram.

We are given

Mass of fat in the soup serving = 5.0 g

Mass of carbohydrates in the soup serving = 20.0 g

Mass of protein in the soup serving = 3.0 g

Conversion factor used:  1 kCal = 1000 Cal

Applying unitary method:

  • <u>For Fat:</u>

1 gram of fat provides 9 calories

So, 5.0 gram of fat will provide = \frac{9}{1}\times 5.0=45Cal=0.045kCal

  • <u>For Carbohydrates:</u>

1 gram of carbohydrates provides 4 calories

So, 20.0 gram of carbohydrates will provide =\frac{4}{1}\times 20.0=80Cal=0.080kCal

  • <u>For Proteins:</u>

1 gram of proteins provides 4 calories

So, 3.0 gram of fat will provide = \frac{4}{1}\times 3=12Cal=0.012kCal

Total kilocalories per serving = [0.045 + 0.080 + 0.012] kCal = 0.137 kCal

Hence, the amount of kilocalories contained in the given serving of soup is 0.137 kCal

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Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron fallsfrom the n = 7 to the n
Helga [31]

Answer:

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.

Explanation:

We recall that Eₙ=\frac{-2.18*10^{-18} J}{n^{2} }

since there was transition from n7 to n=4 we will first calculate the change in the energy i.e ΔE

ΔE=E₄-E₇

ΔE=-2.18*10^{-18} J(\frac{1}{4^{2} } -\frac{1}{7^{2} } )

ΔE=-9.1760*10^-20 J

Now:

|ΔE|=Energy of photon=h*v=h*c/λ

λ=h*c/|ΔE|

λ=\frac{6.63*10^{-34}J.s*3.00*10^{8}m/s  }{9.1760*10^{-20} J }

λ=2.1676*10^-6 m

λ=2167.6*10^-9 m

λ=2167.6 nm

The wavelength of light emitted is 2167.6 nm.

5 0
4 years ago
Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine
yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

7 0
3 years ago
During a combustion reaction, 9.00 grams of oxygen reacted with 3.00 grams of CH4.
Monica [59]

Answer:

0.74 grams of methane

Explanation:

The balanced equation of the combustion reaction of methane with oxygen is:

  • CH₄ + 2 O₂ → CO₂ + 2 H₂O

it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.

firstly, we need to calculate the number of moles of both

for CH₄:

number of moles = mass / molar mass = (3.00 g) /  (16.00 g/mol) = 0.1875 mol.

for O₂:

number of moles = mass / molar mass = (9.00 g) /  (32.00 g/mol) = 0.2812 mol.

  • it is clear that O₂ is the limiting reactant and methane will leftover.

using cross multiplication

1 mol of  CH₄ needs → 2 mol of O₂

???  mol of  CH₄  needs → 0.2812 mol of O₂

∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol

so 0.14 mol will react and the remaining CH₄

mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol

now we convert moles into grams

mass of CH₄ left over = no. of mol of CH₄ left over *  molar mass

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So, the right choice is 0.74 grams of methane

3 0
3 years ago
Which electron configuration represents the electrons of an atom in an excited state? * 2-3 2-7-1 2-8 2-8-1​
Lemur [1.5K]

2-7-1

<h3>Further explanation </h3>

Electrons can move the shell up or down by releasing energy or absorbing energy  

Excited electrons show higher electron transfer to the shell by absorbing energy  

So it can be concluded that there are 2 conditions:  

Ground state is the state of electrons filling shell with the lowest energy levels.  

Excited state is the state of electrons which occupies a higher energy level  

The state of excited electrons can be seen from the presence of electrons which do not fill the skin completely but fill the skin afterward  

2-7-1

From its 8 electron configuration, filling 3 shells, 2 electrons in the firs shell, 7 electrons in the second shell and 1 electron in the third shell

the electrons in the third shell should fill the electrons in the second shell first according to Aufbau rule (lower energy shells)

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8 0
3 years ago
What does heat mean in chemistry
Tpy6a [65]

Answer:

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Explanation:

8 0
3 years ago
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