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Colt1911 [192]
2 years ago
13

How is the bent geometry related to the trigonal planar geometry?

Chemistry
1 answer:
Lorico [155]2 years ago
7 0

Answer:

Molecules with three electron pairs

Explanation:

=>

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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

4 0
3 years ago
Thinner in the center that at the edges; virtual image smaller.
ipn [44]
<h3><u>Answer</u>;</h3>

Concave Lenses

<h3><u>Explanation</u>;</h3>
  • A concave lens is thin in the middle and thick at the edges, such that it seems to cave inwards. It spreads light rays apart producing an image smaller than the actual object.
  • <em><u>Images formed by a concave lens are virtual, upright, reduced in size and located on the same side of the lens as the object. Diverging lenses or concave lens always produce images that share these characteristics. The location of the object does not affect the characteristics of the image. </u></em>
3 0
3 years ago
Read 2 more answers
C9H20 +<br> 02 - &gt; CO2 +<br> H2O<br><br> what is the balanced equation of this
san4es73 [151]

Hey there!

C₉H₂O + O₂ → CO₂ + H₂O

First let's balance the C.

There's 9 on the left and 1 on the right. So, let's add a coefficient of 9 in front of CO₂.

C₉H₂O + O₂ → 9CO₂ + H₂O

Next let's balance the H.

There's 2 on the left and 2 on the right. This means it's already balanced.

C₉H₂O + O₂ → 9CO₂ + H₂O

Lastly, let's balance the O.

There's 3 on the left and 19 on the right. So, let's add a coefficient of 9 in front of O₂.

C₉H₂O + 9O₂ → 9CO₂ + H₂O

This is our final balanced equation.

Hope this helps!

8 0
3 years ago
What is the density of an object that has a mass of 1.08 kg and displaces 50.50 cm3 of water? (Please report your answer in g/cm
masha68 [24]

Answer:

Explanation:

1.08kg/50.5 cm^3(1000g/kg)=21.386 g/cm^3

3 0
2 years ago
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Refer to the following illustration. How many hydrogen atoms could bond with oxygen?
WITCHER [35]
2. 

The reason why is because there are 2 available electron spots on the orbital for the oxygen atom. Hydrogen atoms have one proton and one electron, thus, in order to fill the oxygen atom orbital to a full outer shell, a maximum of 2 atoms could bond with the oxygen atoms. 

6 electrons (oxygen)+ 1 electron (hydrogen)+ 1 electron (hydrogen)= 8 
4 0
3 years ago
Read 2 more answers
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