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slamgirl [31]
2 years ago
11

Household ammonia used for cleaning contains about 10 g (two significant figures) of NH3 in 100 mL (two significant figures) of

solution. What is the molarity of the NH3 in the solution?
Chemistry
1 answer:
lidiya [134]2 years ago
4 0

Answer:

11.44

Explanation:

= Base ionization constant of ammonia =  

m = Mass of ammonia = 5 g

V = Volume of ammonia = 709 mL =  

M = Molar mass of ammonia = 17.03 g/mol

Molaritiy of ammonia

We have the relation

The pH of the solution is 11.44.

Explanation:

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Naily [24]

Answer:

yes I think that they are correct

8 0
3 years ago
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

4 0
3 years ago
40 Points!!! Match the correct definition with the correct keyword.
My name is Ann [436]

Answer:

Okay the Answers are top to bottom.

5, 4, 1, 3, 2

Explanation:

8 0
3 years ago
Choose the element described by the following electron configuration.
nadezda [96]

Answer:

The answer to your question is Argon

Explanation:

Electron configuration given               1s² 2s² 2p⁶ 3s² 3p⁶

To find the element whose electron configuration is given, we can do it by two methods.

Number 1. Sum all the exponents the result will give you the atomic number of the element.

                      2 + 2 + 6 + 2 + 6 = 18

The element with an atomic number of 18 is Argon.

Number 2. Look at the last terms of the electronic configuration

                      3s² 3p⁶

Number three indicates that this element is in the third period in the periodic table.

Sum the exponents    2 + 6 = 8

Number 8 indicates that this element is the number 8 of that period without considering the transition elements.

The element with these characteristics is Argon.

4 0
3 years ago
Read 2 more answers
Please answer the question
Tom [10]

Answer:

Stamens

Explanation:

7 0
2 years ago
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