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2 years ago

Which of the following would have the highest entropy?

Chemistry

1 answer:

ANEK [815]2 years ago

7 0

Answer:

I think it either C. A house made of blocks or D. A pile of laundry.

I hope this helps you :)

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If the human body were a car, glucose would be:

NeTakaya

<span>If the human body were a car, glucose would be the gasoline.
Glucose gives humans energy, we basically run on glucose, among other things, the same way a car would run on gas.
</span>

8 0

2 years ago

The Safe Drinking Water Act (SDWA) sets a limit for mercury-a toxin to the central nervous system-at 0.002 mg/L. Water suppliers

posledela

Answer:

The volume of mercury-contaminated water that has to be consumed to ingest 0.100 g mercury is 2.50 × 10⁴ l

Explanation:

Hi there!

First, let´s convert 0.100 g to mg:

0.100 g · (1000 mg/1 g) = 100 mg

The contaminated water has 0.004 mg per liter, then, we have to find the volume of water that contains 100 mg of mercury:

100 mg · (1 l / 0.004 mg) = 2.50 × 10⁴ l

Then, the volume of mercury-contaminated water ( at a concentration of 0.004 mg/l) that has to be consumed to ingest 0.100 g mercury is 2.50 × 10⁴ l

Have a nice day!

8 0

2 years ago

A substance is 89.2% carbon by mass. how much of the substance would be needed to recover 34.6 mol of pure carbon?

kari74 [83]

1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon,
then, (415.2/89.2) ×100 = 465.47 g of the substance will be required to yield 34.6 moles of carbon.

8 0

2 years ago

Read 2 more answers

When balanced, which equation would have the coefficient 3 in front of any of the reactants? Zn + HCl ZnCl2 + H2 H2SO4 + B(OH)3

Gwar [14]

The correct answer is B. H2SO4 + B(OH)3 B2(SO4)3 + H2O

Hope this helps!

6 0

2 years ago

Read 2 more answers

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

2 years ago