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ELEN [110]
3 years ago
5

The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

reviation (w/v) is also common. How many grams of sucrose are needed to make 765 mL of a 31.0 % (w/v) sucrose solution?
Chemistry
1 answer:
ololo11 [35]3 years ago
4 0

Answer:

There are needed 237.15 g of sucrose.

Explanation:

The 31.0 % (w/v) of the sucrose solution means that in 100 ml of solution, you have 31 g of solute, in this case sucrose.

So you want to make a solution with 765 mL.

Let's think the rule of three:

100 mL solution__contain __31 g sucrose

765 mL solution _________ (765 . 31 ) /100 = 237.15 g

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jekas [21]

Answer:

The number of tires required to power ten homes for one year is approximately 2,730 tires

Explanation:

The given information are;

The energy generated by burning one rubber tire = 250,000 BTUs

1 BTU = 0.0003 kW

The amount of energy consumed by the average American home = 10,000 kWh

Therefore, the amount of energy generated by burning one tire in kW can be found as follows;

The energy generated by burning one rubber tire = 250,000 BTUs

1 BTU = 0.0003 kW

∴ 1 BTU × 250,000 = 250,000 BTUs = 0.0003 kW × 250,000 = 75 kW

We note that 250,000 BTUs is equivalent to 263764 kJ

1 kWh is equivalent to 3600 kJ

10,000 kWh = 10,000 × 3600 = 36,000,000 kJ

The energy requirement for ten homes  for one year = 10 × 36,000,000 kJ = 360,000,000 kJ

At 50% efficiency, the energy produced per tire = 0.5 × 263764 kJ = 131,882 kJ

The electrical energy produced per tire = 131,882 kJ/tire

The number of tires required to power ten homes for one year = 360,000,000 kJ/  131,882 kJ/tire = 2,729.71293  ≈2,730 tires

The number of tires required to power ten homes  ≈2,730 tires.

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