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3 years ago

A _____ exists in the same physical state as that of the reaction it catalyzes.

1 answer:

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A Homogeneous Catalyst exists in the same physical state as that of the reaction it catalyzes.

Explanation:
Catalysts are classifies as follow,
1. Homogeneous Catalyst
2. Heterogeneous Catalyst

Homogeneous Catalyst are those which are in same phase as that of reactants are.

Example:

Acetic Acid + Ethanol + H₂SO₄ → Ethyl Acetate + H₂O + H₂SO₄

In this reaction catalyst and reactants all are in same phase i.e. Liquids.

Heterogeneous Catalyst are those which are in different phase as that of reactants are.

Example:

Ethene + H₂ + Ni → Ethane + Ni

In this reaction ethene and hydrogen are gases while Nickle catalyst is in solid state.

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To measure the amount of nickel in some industrial waste fluid, an analytical chemist adds sodium hydroxide solution to a sample

Cloud [144]

Answer:

The final balanced equation is

Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+

Explanation:

It is given that sodium hydroxide is added to collect the solid nickel(II) hydroxide product

The empirical equation for this statement is

Ni2+ + NaOH --> Ni (OH)2 + Na+

We will first balance the hydroxide molecule. On the right side there are two OH molecules.

Thus, on the left side we will take 2 sodium hydroxide

Ni2+ + 2NaOH --> Ni (OH)2 + Na+

Now we will balance the sodium ion which are 2 in numbers on the left side and 1 on the right side

Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+

So, the final balanced equation is

Ni2+ + 2NaOH --> Ni (OH)2 + 2Na+

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3 years ago

How do we find mass number of elements in chemistry ?

sukhopar [10]

Answer: Atomic number, atomic mass, and relative atomic mass

Explanation:

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3 years ago

Carson’s class visited a hydroelectric plant on a field trip during their study of energy. The process interested him, and he ha

AURORKA [14]

Answer:

look in your science book

Explanation:

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2 years ago

What is the percent of O in CO2?

katen-ka-za [31]

Answer:

72.7% of oxygen

Explanation:

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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