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Georgia [21]
3 years ago
11

A compound is found to contain 58.80 % xenon, 7.166 % oxygen, and 34.04 % fluorine by mass. what is the empirical formula for th

is compound?
Chemistry
1 answer:
marissa [1.9K]3 years ago
6 0
To make it easier, assume that we have a total of 100 g of a compound. Hence, we have 58.80g of xenon, 7.166g of oxygen, and 34.04g of fluorine. 
Know we will convert each of these masses to moles by using the atomic masses:

58.8/131.3 = 0.45 mole of Xe
7.166/16 = 0.45 mole of O
34.04/19 = 1.79 mole of F

Now, we will divide all the mole numbers by the smallest among them and get the number of atoms in the compound:

Xe = 0.45/0.45 = 1
O = 045/0.45 = 1
F = 1.79/0.45 = 3.98 = 4

So, the empirical formula of the compound XeOF₄
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Answer: The empirical formula for C6H12O6 is CH2O. Every carbohydrate, be it simple or complex, has an empirical formula CH2O

Explanation:

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Which regions on the periodic table can adopt positive and negative oxidation numbers?
Dennis_Churaev [7]

<span>The region(s) of the periodic table which are made up of elements that can adopt both positive and negative oxidation numbers are the “non-metal” region. As we can see on the periodic table, the elements situated at the right side of the table have two oxidation states, one positive and the other a negative. </span>

7 0
3 years ago
Which of the following is a buffer system? Which of the following is a buffer system? H2CO3(aq) and KHCO3(aq) NaCl(aq) and NaOH(
Readme [11.4K]

Answer:

Explanation:

A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.

In the systems:

H₂CO₃(aq) and KHCO₃(aq): Carbonic acid, H₂CO₃, is a weak acid that, in solution with its conjugate pair, HCO₃⁻ make a <em>buffer system.</em>

NaCl(aq) and NaOH(aq): NaCl is a salt and NaOH is a strong base. Thus, this system <em>is not </em> a buffer system.

H₂O(l) and HCl(aq): Water is a solvent and HCl a strong acid. This <em>is not </em>a buffer system.

HCl(aq) and NaOH(aq): HCl is a strong acid and NaOH a strong base. This <em>is not </em>a buffer system.

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3 0
3 years ago
What is the molarity of solution that is 5.50 percentage by mass oxalic acid and has a density of 1.024 g/ml
Y_Kistochka [10]

Answer:

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.

Explanation:

Mass percentage of oxalic acid = 5.50%

This means that in 100 grams of solution there are 5.50 grams of oxalic acid.

Mass of solution , m = 100

Volume of the solution = V

Density of the solution = d = 1.024 g/mL

V=\frac{m}{d}=\frac{100 g}{1.024 g/mL}=97.66mL

V = 97.66 mL = 0.09766 L

(1 mL = 0.001 L)

Moles of oxalic acid = \frac{5.50 g}{90 g/mol}=0.06111 mol

Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}

The molarity of the solution :

=\frac{0.06111 mol}{0.09766  L}=0.6257M

0.6257 M is the molarity of solution that is 5.50 percentage by mass oxalic acid.

6 0
3 years ago
Winter is coming and it's time to make sure you have the right amount of antifreeze
ololo11 [35]

The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of  ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

Required

The molality

Solution

molality = mol of solute/ 1 kg solvent

mol of solute = mol of ethylene glycol

  • mass of ethylene glycol :

= volume x density

= 6 L x 1.1132 kg/L

= 6.6792 kg

= 6679.2 g

  • mol of ethylene glycol (MW=62.07 g/mol)

=mass : MW

=6679.2 : 62.07

=107.608

  • mass of water

6 L water = 6 kg water(ρ= 1 kg/L)

  • molality

\tt =\dfrac{107.608}{6}=17.93~m

5 0
3 years ago
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