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2 years ago

QUESTION 2

Chemistry

1 answer:

Aleksandr [31]2 years ago

4 0

Answer:

you can see the answer at the pic

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What is the process of breaking down food so that it can be used by living things??

Darya [45]

I think the answer you're looking for is digestion.

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2 years ago

James decides to walk home from school today. He lives 3 miles from school and can walk home in 45 minutes. At what rate is Jame

Natasha2012 [34]

One mile every 15 minutes

7 0

3 years ago

Calculate the mass of H2OH2O produced by metabolism of 2.4 kgkg of fat, assuming the fat consists entirely of tristearin (C57H11

otez555 [7]

Answer:

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

Explanation:

$2C_{57}H_{110}O_6+163O_2\rightarrow 114CO_2+110H_2O$

Mass of fat = 2.4 kg = 2.4 × 1000 g = 2400 g

1 kg = 1000 g

Molar mass of fat = M

M = 57 × 12 g/mol + 110 × 1 g/mol+ 6 × 16 g/mol = 890 g/mol[/tex]

Moles of fat = $\frac{2400 g}{890 g/mol}=2.6966 mol$

According to reaction , 2 moles of fat gives 110 moles of water. Then 2.6966 moles of fat will give ;

$\frac{110}{2}\times 2.6966 mol=148.31 mol$ of water

Mass of 148.31 moles of water ;

148.31 mol × 18 g/mol = 2,669.58 g

2,669.58 grams of water will be produced by metabolism of 2.4 kilogram of fat.

8 0

2 years ago

140000 rounded to 3 sig gigs

attashe74 [19]

1.40E+5 (140000)
I believe this is right

8 0

2 years ago

A student wants to make a 0.150 M aqueous solution of silver (I) nitrate but only has 11.27 g of AgNO3. What volume of the 0.150

Usimov [2.4K]

Answer:

442.3 mL

Explanation:

Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

$M=\frac{Moles of substance X}{Volume of the solution}$

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

$11.27 g AgNO_{3} *\frac{1 mole AgNO_{3}}{169.87 g AgNO_{3}} = 0.06634 moles AgNO_{3}$

Then you can solve for the volume of the solution:

$Volume of the solution=\frac{Moles of AgNO_{3}}{M} =\frac{0.06634 mol AgNO_{3}}{0.150 M} =0.4423 L = 442.3 mL$

Hope it helps!

3 0

3 years ago