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2 years ago

A plastic wading pool has a circumference of 12.56 feet. Which procedure will help you

Chemistry

1 answer:

RoseWind [281]2 years ago

7 0

The diameter of the pool is 4

Explanation:

To calculate the diameter we must write the values given in the question, here

The circumference of the pool is given as C=12.56 feet.

To calculate the diameter of the pool,

If radius is given, Multiply the radius by 2

If Circumference is given divide the circumference by Pi (π)

the constant value for Pi (π)=3.14

Diameter(D)= C/π

12.56/3.14

The diameter for the plastic wading pool is 4

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What kind of charge does a neutron have?

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Neutrons actually don't carry an electrical charge, which is why they are called neutrons because they are "Neutral".

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3 years ago

Density is d= m/vol. If a material has a mass of 65.5 g and a volume of 32.5 ml, it has a density of

ryzh [129]

Using the given formula, the density of the material is 2.015 g/mL

<h3>Calculating Density </h3>

From the question, we are to determine the density of the material

From the given formula

Density = Mass / Volume

And from the given information,

Mass = 65.5 g

and volume = 32.5 mL

Putting the parameters into the equation,

Density = 65.5/32.5

Density = 2.015 g/mL

Hence, the density of the material is 2.015 g/mL.

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2 years ago

To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber

Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

$\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}$

7 0

3 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

Option d: 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0

3 years ago

Please i need help really quickly on this. thank you

prohojiy [21]

Answer:

the error could have been the fact that the unit for volume wasn't changed from cm³ to dm³

hence the calculation error

the solution to this would be first dividing the volume by 1000 to get that same amount in dm³ which is the standard unit to be used for volume-density calculations

6 0

3 years ago