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Fantom [35]
3 years ago
14

Which of the following has to remain constant to allow you to use the combined gas law for calculations?

Chemistry
1 answer:
Mazyrski [523]3 years ago
4 0

Answer:

Amount of gas

Explanation:

The combined gas law has to do with pressure, volume, and temperature. All of them can be a variable that you have to solve for. However, amount of gas is not a variable you can solve for, which means it must remain constant.

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A cylinder is labeled \"PENTANE.\" When the gas inside the cylinder is monochlorinated, five isomers of formula C5H11Cl result.
melisa1 [442]

Answer:

a mixture of two these

Explanation:

The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.

n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.

Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.

Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.

Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.  

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2 years ago
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How is Earth’s outer layer different from a cracked hard-boiled egg?
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3 years ago
The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate
Delicious77 [7]

<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

<u>For b:</u> The atomic radius of potassium is 231.01 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the lattice parameter or edge length of the crystal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 0.855g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal = 39.09 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell = ?

Putting values in above equation, we get:

0.855=\frac{2\times 39.09}{6.022\times 10^{23}\times (a)^3}\\\\\a=5.335\times 10^{-8}cm=533.5pm

<u>Conversion factor:</u>  1cm=10^{10}pm

Hence, the edge length of the crystal is 533.5 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 533.5 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 533.5}{4}=231.01pm

Hence, the atomic radius of potassium is 231.01 pm

4 0
3 years ago
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