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3 years ago

15

What is a real life example of Charles’ law?

1 answer:

max2010maxim [7]3 years ago

6 0

Leaving a basketball out in the cold weather. When a basketball if left in a cold garage or outside during the cold months, it loses its air inside (or volume).

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Our solar system formed from a

ankoles [38]

Our solar system formed about 4.5 billion years ago from a dense cloud of interstellar gas and dust. The cloud collapsed most likely due to a shockwave form a nearby exploding start called a supernova

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3 years ago

PLEASE HELP <br> I do not understand

Vikentia [17]

Answer:

Absorbing beta particle because the beta is the numbers and are less and the big numbers are positive and they are the alpha so when you add beta particle it is called Absorbing so the answer is Absorbing beta particle

4 0

2 years ago

Read 2 more answers

AURORKA [14]

Answer:

hi

Explanation:

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4 0

3 years ago

The density of a 45.0 mass % solution of ethanol (C2H5OH) in water is 0.873 g/mL. What is the molarity of the solution?

antoniya [11.8K]

Answer : The concentration of solution is, 8.53 M.

Explanation :

As we are given, 45.0 mass % solution of ethanol in water that means 45.0 g of ethanol present in 100 g of solution.

First we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}$

$\text{Volume of solution}=\frac{100g}{0.873g/mL}=114.5mL$

Now we have to calculate the molarity of solution.

Mass of $C_2H_5OH$ = 45.0 g

Volume of solution = 114.5 mL

Molar mass of $C_2H_5OH$ = 46.07 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of }C_2H_5OH\times 1000}{\text{Molar mass of }C_2H_5OH\times \text{Volume of solution (in mL)}}$

Now put all the given values in this formula, we get:

$\text{Molarity}=\frac{45.0g\times 1000}{46.07g/mole\times 114.5mL}=8.53mole/L=8.53M$

Therefore, the concentration of solution is, 8.53 M.

6 0

3 years ago

How many seconds does it take a race horse to run six furlongs at 40.7 miles per hour if: I furlong = 40 rods; 5.5 yards = 1 rod

lana [24]

Distance traveled by the race horse=6 furlongs

Converting the distance from furlongs to rods: 40 rods =1 furlong

$6 furlongs*\frac{40rods}{1furlong} =240rods$

Converting the distance from rods to yards, feet and miles: 5.5 yards = 1 rod, 3foot =1 yard, 1 mile = 5280 feet.

$240rods*\frac{5.5yards}{1rod}*\frac{3foot}{1yard}*\frac{1mi}{5280feet}=0.75mi$

The given speed of race horse = 40.7mi/hr

Calculating the time required:

$0.75mi*\frac{1hr}{40.7mi}*\frac{60min}{1hr}*\frac{60s}{1min}= 66.34s$

Therefore, 66.34 s is required for a race horse to run six furlongs at 40.7 miles per hour.

8 0

3 years ago