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arsen [322]
3 years ago
5

Why would an electrician use parallel circuits to wire the lights in each room of a house instead of using a series circuit?

Chemistry
1 answer:
aivan3 [116]3 years ago
5 0
B.
When electricians get their license, they are only licensed to install parallel circuits. Most electricians aren't certified to install series circuits.
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Design an invitation you could use to calculate the density of a penny
Nataliya [291]
Firstly, the density of any substance is represented by the mass (amount of matter) as divided by the volume(amount of space). According to external websites, the mass of a penny is 2.5 grams.However, the volume of a penny is .35cm to the power of 3 (due to the thickness of the penny being extremely minimal.Thus the amount of density is extremely little). Therefore, the density of a penny is 0.875 g/cm cubed (dimensional analysis).As for an invention that could be used, that is possible with the usage of a series of measurements that can both calculate mass and volume and directly allocate that to attain density
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4 years ago
What does the letter "E" represents when in a chemical equation
Amiraneli [1.4K]

Answer:

1. Equality

<u><em>Hope this helps :)</em></u>

<em><u>Pls brainliest...</u></em>

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2 years ago
How do you make a model of something you cannot see?
hoa [83]

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Explanation:

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3 0
2 years ago
Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl
Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = V

V_2 = final volume of gas = 2V

n_1 = initial moles of gas  = 1.56 mole

n_2 = final moles of gas  = ?

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

\frac{V}{2V}=\frac{1.56mole}{n_2}

n_2=3.12mole

Now we have to calculate the mass of helium gas at doubled volume.

\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He

\text{Mass of }He=3.12mole\times 4g/mole=12.48g

Therefore, the mass of helium gas added must be 12.48 grams.

4 0
3 years ago
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