Chemistry is not my subject please help

If the length, width, and height of a box are 10.00 cm, 7.25 cm and 3.00 cm, respectively, what is the volume of the box in unit

Blizzard [7]

Answer:

For a: The volume of the box is 217.5 mL

For b: The volume of the box is 0.2175 L

Explanation:

The box is a type of cuboid.

To calculate the volume of cuboid, we use the equation:

$V=lbh$

where,

V = volume of cuboid

l = length of cuboid = 10.00 cm

b = breadth of cuboid = 7.25 cm

h = height of cuboid = 3.00 cm

Putting values in above equation, we get:

$V=10.00\times 7.25\times 3.00=217.5cm^3$

For a:

To convert the volume of cuboid into milliliters, we use the conversion factor:

$1mL=1cm^3$

So,

$\Rightarrow 217.5cm^3\times (\frac{1mL}{1cm^3})\\\\\Rightarrow 217.5mL$

Hence, the volume of the box is 217.5 mL

For b:

To convert the volume of cuboid into liters, we use the conversion factor:

$1L=1000cm^3$

So,

$\Rightarrow 217.5cm^3\times (\frac{1L}{1000cm^3})\\\\\Rightarrow 0.2175L$

Hence, the volume of the box is 0.2175 L

3 0

3 years ago

Jordan needs to repeat the experiment but his teacher says that he needs to improve his design in his second experiment what sho

Margarita [4]

Just add more detail in the second experiment explain every little thing.

8 0

2 years ago

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Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

givi [52]

Answer: The amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

Explanation:

We are given:

Moles of electron = 1 mole

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ number of particles.

We know that:

Charge on 1 electron = $1.6\times 10^{-19}C$

Charge on 1 mole of electrons = $1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C$

$Cu^{2+}+2e^-\rightarrow Cu$

$2\times 96500=193000C$ is passed to deposit = 1 mole of copper

63.5 g of copper is deposited by = 193000 C

$14\times 1000g=14000g$ of copper is deposited by = $\frac{193000}{63.5}\times 14000=42551181 C$

To calculate the time required, we use the equation:

$I=\frac{q}{t}$

where,

I = current passed = 40.0 A

q = total charge = 42551181 C

t = time required = ?

Putting values in above equation, we get:

$40.0=\frac{42551181 C}{t}\\\\t=1063779sec$

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, $1063779s\times \frac{1hr}{3600s}=295hr$

Hence, the amount of time needed to plate 14.0 kg of copper onto the cathode is 295 hours

3 0

3 years ago

How many grams of NaBr are needed to produce 187 grams of CaBr2

andrey2020 [161]

You will need 96.3g (3sf)

6 0

3 years ago

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During a long drought, many deer become weak because a basic need is not being met. What requirements for life are the deer most

saw5 [17]

Enough water, I actually took this test! <3

- S0ft1e

3 0

2 years ago

Read 2 more answers