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3 years ago

Help me if you know this pleaseeee

Chemistry

2 answers:

yKpoI14uk [10]3 years ago

7 0

Answer:

help you with what?

Explanation:

azamat3 years ago

3 0

Answer:

What is the question?

Explanation:

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Which is an example of a compound?

makkiz [27]

Answer:

Octane - Formula: C8H^18 = Carbon^8 + Hydrogen^18

Explanation: Octane is a compound because there are 8 atoms of carbon and 18 atoms of hydrogen in one molecule of C8H18. There are also 8 moles of carbon and 18 moles of hydrogen.

5 0

3 years ago

What is the pOH of a 0.0072 M solution of sodium hydroxide? pOH=?

Rus_ich [418]

POH of a 0.0072 M=-lg(0.0072) = 2.1426675

8 0

3 years ago

Read 2 more answers

How many moles of H2O are needed to produce 5.6 mol of NaOH? Na2O + H2O --> 2NaOH

navik [9.2K]

Answer: 2.8 moles

Explanation:

The balanced equation below shows that 1 mole of sodium oxide reacts with 1 mole of water to form 2 moles of sodium hydroxide respectively.

Na2O + H2O --> 2NaOH

1 mole of H2O = 2 moles of NaOH

Let Z moles of H2O = 5.6 mole of NaOH

To get the value of Z, cross multiply

5.6 moles x 1 mole= Z x 2 moles

5.6 = 2Z

Divide both sides by 2

5.6/2 = 2Z/2

2.8 = Z

Thus, 2.8moles of H2O are needed to produce 5.6 mol of NaOH

3 0

3 years ago

A compound of mercury and oxygen is heated in order to decompose the compound. A 4.08 grams sample of mercury oxide upon heating

arlik [135]

Answer:

HgO (empirical formula)

Explanation:

4.08 - 3.78 = 0.3g (oxygen)

$(\frac{4.08}{201}) \: \: \: (\frac{0.3}{16} )$

0.02 : 0.02

0.02/0.02 : 0.02/0.02

1 : 1 (ratio)

HgO ( empirical formula)

2HgO ----> 2Hg + O2 ( your equation correct)

7 0

2 years ago

A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

Step 1: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

Step 3: Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

Step 4: Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

Step 5: Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

Step 6: Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

Step 7: Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

Step 8: Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

Step 9: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

Step 10: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0

3 years ago