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mixas84 [53]
3 years ago
6

What are the elements of PCl5 and the name

Chemistry
1 answer:
prisoha [69]3 years ago
8 0
Phosphorus, Chlorine. The IUPAC name is Phosphorus pentachloride

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Why does it take less
KengaRu [80]

Answer:I’m not for sure

Explanation:

8 0
3 years ago
The half-life of plutonium-239 is 24300. If a nuclear bomb released 8kg of this isotope, how many years would pass before the am
Aneli [31]

Answer:

= 72900 years

Explanation:

  • The half-life is the time taken by a radioactive material to decay by half the original amount.
  • The half-life of plutonium-239 is 24300 years which means it takes 24300 years to decay by half the original amount.

To calculate the time taken for a mass of 8 kg to decay to 1 kg we use;

New mass = Original mass x (1/2) ^n, where n is the number of half-lives

Therefore;

1 kg = 8 kg × (1/2)^n

1/8 = (1/2)^n

solving for n;

n =3

Therefore;

Time = 3 × 24300 years

          = 72900 years

It will, therefore, take 72900 years for 8 kg of plutonium-239 to decay to 1 kg.

3 0
3 years ago
Need help !!!!! ASAP
Julli [10]
<h2>Hello!</h2>

The answer is:

The new volume will be 1 L.

V_{2}=1L

<h2>Why?</h2>

To solve the problem, since we are given the volume and the first and the second pressure, to calculate the new volume, we need to assume that the temperature is constant.

To solve this problem, we need to use Boyle's Law. Boyle's Law establishes when the temperature is kept constant, the pressure and the volume will be proportional.

Boyle's Law equation is:

P_{1}V_{1}=P_{2}V_{2}

So, we are given the information:

V_{1}=2L\\P_{1}=50kPa\\P_{2}=100kPa

Then, isolating the new volume and substituting into the equation, we have:

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\frac{P_{1}V_{1}}{P_{2}}

V_{2}=\frac{50kPa*2L}{100kPa}=1L

Hence, the new volume will be 1 L.

V_{2}=1L

Have a nice day!

4 0
3 years ago
How many moles of copper are present in a sample that contains 4.55 x 10^24 atoms of copper
FinnZ [79.3K]

Hope you could understand.

If you have any query, feel free to ask.

3 0
2 years ago
How many grams of chlorine gas are present in a 150. liter cylinder of chlorine held at a pressure of 1.00 atm and 0. °C? Group
OlgaM077 [116]

Answer:

474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

In this case:

  • P= 1.00 atm
  • V= 150 L
  • n= ?
  • R= 0.082 \frac{atm*L}{mol*K}
  • T= 0 C= 273 K

Replacing:

1.00 atm* 150 L= n*0.08206 \frac{atm*L}{mol*K} *273 K

Solving:

n=\frac{1.00 atm* 150 L}{0.08206 \frac{atm*L}{mol*K}*273 K}

n= 6.69 moles

Being Cl= 35.45 g/mole, the molar mass of chlorine gas is:

Cl₂=2*35.45 g/mole= 70.9 g/mole

So if 1 mole has 70.9 grams, 6.69 moles of the gas, how much mass does it have?

mass=\frac{6.69 moles*70.9 grams}{1 mole}

mass= 474.321 grams ≅ 474 grams

<u><em>474 grams of chlorine gas are present in a 150 liter cylinder of chlorine held at a pressure of 1.00 atm and 0 °C</em></u>

4 0
3 years ago
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