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Thepotemich [5.8K]
3 years ago
8

Bird wings and dragonfly wings are both used for flight.

Chemistry
2 answers:
vredina [299]3 years ago
7 0

Answer:

Yes

Explanation:

What's the question?

scoray [572]3 years ago
7 0

Answer:

yes, they both use it to fllight.

Explanation:

As they are adapted to flight to various places to catch their prey and be safe from various type of harms.

<em><u>hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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If 1.80 moles of NaCl was dissolved in enough water to make 3.60 L of solution, what is the Molarity?
sweet-ann [11.9K]

Answer:

0.5M

Explanation:

The equation for molarity is:

  • M = \frac{mol}{liters} ; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.

We are given that there are:

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  • 3.60 Liters of solution (the volume in liters of solution)

Now we just plug those numbers into the formula and get our answer:

  • M= \frac{1.80mol}{3.60L}= 0.5M

After doing the math and dividing the moles of solute by the liters of solution, we get that the molarity of the solution is 0.5M.

8 0
3 years ago
How does knowing the equivalence point for a titration help pick out an indicator?
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Calculate the number of grams of solute needed to make each of the following solutions:
AleksAgata [21]

Explanation:

(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

1) 100 g of 0.500% (w/w) NaI

Mass of solution = 100 g

Mass of solute = x

Required w/w % of solution = 0.500%

0.500\%=\frac{x}{100 g}\times 100

x=\frac{0.500\times 100 g}{100}=0.500 g

0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.

2) 250 g of 0.500% (w/w) NaBr

Mass of solution = 250 g

Mass of solute = x

Required w/w % of solution = 0.500%

0.500\%=\frac{x}{250 g}\times 100

x=\frac{0.500\times 250 g}{100}=1.25 g

1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr

3) 500 g of 1.25% (w/w) glucose

Mass of solution = 500 g

Mass of solute = x

Required w/w % of solution = 1.25%

1.25\%=\frac{x}{500 g}\times 100

x=\frac{1.25\times 500 g}{100}=6.25 g

6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)

4) 750 g of 2.00% (w/w) sulfuric acid.

Mass of solution = 750 g

Mass of solute = x

Required w/w % of solution = 2.00%

2.00\%=\frac{x}{750 g}\times 100

x=\frac{2.00\times 750 g}{100}=15.0 g

15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.

3 0
3 years ago
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