In the chemical formula AuCI3, determine the elements and how many atoms of each one are present

Melting point is dependent on the intermolecular forces which means the bonds between the molecules of bromine as it is a simple molecular structure the intermolecular bonds of bromine are weak bcz they are weak vandervaal forces thats why Bromines melting point is low..In short when intermolecular bomds are weak the M.P is lower

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2 years ago

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Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume

pishuonlain [190]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

mass = 0.508 g, Volume = 0.175 L

Temperature = (25 + 273) K = 298 K, P = 1 atm

As per the ideal gas law, PV = nRT.

where, n = no. of moles = $\frac{mass}{\text{molar mass}}$

Hence, putting all the given values into the ideal gas equation as follows.

PV = $\frac{mass}{\text{molar mass}} \times RT$

1 atm \times 0.175 L = $\frac{0.508 g}{\text{molar mass}} \times 0.0821 L atm/ K mol \times 298 K$

= 71.02 g

As the molar mass of a chlorine atom is 35.4 g/mol and it exists as a gas. So, molar mass of $Cl_{2}$ is 70.8 g/mol or 71 g/mol (approx).

Thus, we can conclude that the gas is most likely chlorine.

4 0

2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.