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finlep [7]
2 years ago
9

A rectangular aquarium is 47.2 cm wide, 89 cm long, and 65.38 cm tall. What is the volume of the aquarium? How many liters of wa

ter does it take to fill this aquarium?
Chemistry
1 answer:
maria [59]2 years ago
6 0

Answer: 274.6 L

Explanation:

Given : length = 89 cm, width = 47.2 cm and height = 65.38 cm

Volume of rectangular aquarium =length\times width\times height

Volume of aquarium =89cm\times 47.2cm\times 65.38cm=274648.3cm^3=274.6L        (1000cm^3=1L)

Thus the tank can hold 274.6 L of water to fill this aquarium.



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3 years ago
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Time that is measured in definite periods such as minutes, day,and years is called
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7 0
3 years ago
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

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