All categories

2 years ago

A 1.19 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm. What is the identity of the halogen

Chemistry

1 answer:

Dimas [21]2 years ago

7 0

Answer:

The halogen is Iodine.

Explanation:

Using the ideal gas equation, we find the number of moles of gas present, n.

PV = nRT where P = pressure of gas = 1.41 atm, V = volume of gas = 109 mL = 0.109 L, n = number of moles of gas, R = molar gas constant = 0.082 L-atm/mol-K and T = temperature of gas = 398 K

Since PV = nRT, making n subject of the formula, we have

n = PV/RT

substituting the values of the variables into the equation, we have

n = 1.41 atm × 0.109 L/(0.082 L-atm/mol-K × 398 K)

n = 0.15369 atm-L/32.636 L-atm/mol

n = 0.0047 mol

Since n = m/M where m = mass of gas = 1.19 g and M = relative molecular mass of gas

So, M = m/n

M = 1.19 g/0.0047 mol

M = 252.7 g

Since halogens are diatomic the relative atomic mass is M/2 = 252.7g/2 = 126.34 g

From tables, the only halogen with this atomic mass is Iodine.

So, the halogen is Iodine.

You might be interested in

What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

Round

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.4184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately 147 grams.

3 0

3 years ago

9. How many grams are in 2.0 x 105 molecules of carbon monoxide?

timama [110]

Answer: 9.3 x 10^ 18 g CO

Explanation:

Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:

(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO

This cancels molecules CO.

Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:

(1 mol CO) x 28 g CO / 1 mol CO

This cancels mol CO, which leaves grams CO.

5 0

3 years ago

Why does 10g Nitrogen gas contain the same number of molecules as 10 g of carbon monoxide

icang [17]

Answer:

$\fbox\colorbox{cyan}{❥i \: hope \: it \: help \: you}$