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Mice21 [21]
3 years ago
9

A sample of potassium has an average atomic mass of 39.0983amu. There are three isotopic forms of potassium element in the sampl

e. About 93.2581% of the potassium atoms are 38.9637amu mass (39K); 0.0117% are 39.9639amu mass (40K), and the remaining isotope 41K is of 40.9618amu mass. Calculate the percentage composition of 41K. Report your answer with five significant figures.
Chemistry
1 answer:
igomit [66]3 years ago
7 0

Answer:

Percent Composition of 41K = 6.7302%

Explanation:

The explination is in the image.

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Plants contain xylem and phloem tissues for carrying water and food. What organ system in animals performs a similar function as
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How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben
Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(C) 17 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L  =   1000 mL

1:200 solution implies the \frac{weight}{volume} in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be \frac{1000mL}{200mL}=\frac{1g}{xg}

200mL × 1g = 1000 mL × x(g)

x(g) = \frac{200mL*1g}{1000mL}

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ \frac{10mL}{250mL}=\frac{0.2g}{y(g)}

y(g) = \frac{250mL*0.2g}{10mL}

y(g) = 5g of benzalkonium chloride.

Now, at 17% \frac{weight}{volume} concentrate contains 17g/100ml:

∴  the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= \frac{17g}{5g} = \frac{100mL}{z(mL)}

z(mL) = \frac{100mL*5g}{17g}

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

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What would you call the following chemical reaction, Zn(s) + H2O(g) – Zno (s) + H2(g)
egoroff_w [7]

I'd say it's single replacement/displacement

5 0
3 years ago
80 protons, 119 neutron
g100num [7]
Mercury-199 is composed of 80 protons, 119 neutrons, and 80 electrons.
4 0
2 years ago
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