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Westkost [7]
3 years ago
7

Percentage of isotopes in 35/17 Cl and 37/17 Cl with the molar mass of chlorine as 35.5g and the mass of each nucleon to be 1.00

00 amu
Chemistry
1 answer:
AlekseyPX3 years ago
4 0

Answer:

The abundance of 35/17 Cl is 75% and 25% of 37/17Cl

Explanation:

The molar mass of an atom is defined as the sum of the masses of each isotope times its abundance, that is:

35.5amu = 35amu*X + 37amu*Y <em>(1)</em>

<em>Where X is the abundance of the 35/17Cl and Y the abundance of 37/17 Cl</em>

<em />

As there are just these 2 isotopes and the abundances of both isotopes = 1:

X + Y = 1 <em>(2)</em>

<em></em>

Replacing (2) in (1):

35.5 = 35X + 37(1-X)

35.5 = 35X + 37 - 37X

-1.5 = -2X

X = 0.75 = 75%

And Y = 100-75% = 25%

<h3>The abundance of 35/17 Cl is 75% and 25% of 37/17Cl</h3>
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The difference in an area with high concentration and an area with low concentration is called the concentration gradient.

<h3>What is Concentration Gradient ?</h3>

A concentration gradient occurs when the concentration of particles is higher in one area than another.

In passive transport, particles will diffuse down a concentration gradient, from areas of higher concentration to areas of lower concentration, until they are evenly spaced.

This difference in an area with high concentration and an area with low concentration is called the concentration gradient.

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1 year ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
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Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

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2 years ago
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The serving of peanut butter contains 117kcal

<h3>Calculation of calories of food Nutrients</h3>

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1 gram of fat contains 9 (kcal)

But peanut butter contains 5g of fat. The kilocalories of fat present is;

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