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3 years ago

I really need help with this ASAP

Chemistry

1 answer:

Norma-Jean [14]3 years ago

6 0

Answer:

1) Reaction rate, in chemistry, the speed at which a chemical reaction proceeds. It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time.

2) By collecting the hydrogen gas that is produced over water or in a syringe, rate graphs can be produced. The volume of gas produced and the time taken need to be recorded. The rate of the same reaction could be monitored by measuring the change in the mass of reactants as they react to form products.

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Under the conditions of 775 torr of pressure and 30 °C temperature,

docker41 [41]

Answer:

31 ml

Explanation:

At constant temperature of 30°C;

Using Boyles Law:

$P1V1 = P2V2$

775 × 30 = 750V2

V2 = (775 × 30) / 750 = 31ml

6 0

3 years ago

Nickel metal will react with CO gas to form a compound called nickel tetracarbonyl (Ni(CO)4), which is a gas at temperatures abo

Bad White [126]

Answer:

The final total pressure in the bulb will be 0.567 atm.

Explanation:

The equation of the reaction is:

Ni + 4CO → Ni(CO)₄

The pressure in the bulb will be the sum of the pressures of each gas (remaining CO and Ni(CO)₄ produced).

The pressure of each gas can be calculated using this equation:

For the gas Ni(CO)₄:

P(Ni(CO)₄) = n * R * T / V

where:

P(Ni(CO)₄) = pressure of Ni(CO)₄

n = number of moles of Ni(CO)₄.

R = gas constant = 0.082 l amt / K mol

T = temperature

V = volume

So we have to find how many moles of Ni(CO)₄ were produced and how many moles of CO remained unreacted.

We can calculate the initial number of moles of CO with the data provided in the problem:

P(CO) = n * R * T / V

solving for n:

P(CO) * V / R * T = n

Replacing with the data:

1.20 atm * 1.50 l / 0.082 (l atm / K mol) * 346K = n

n = 0.06mol.

Now we know how many moles of CO were initially present.

To know how many moles of Ni(CO)₄ were produced, we have to find how many Ni reacted with CO.

Initially, we have 0.5869 g of Ni, which is (0.5869 g * 1 mol/58.69 g) 0.01 mol Ni.

From the chemical equation, we know that 1 mol Ni reacts with 4 mol CO, therefore, 0.01 mol Ni will react with 0.04 mol CO producing 0.01 mol Ni(CO)₄ (see the chemical equation above).

At the end of the reaction, we will have 0.01 mol Ni(CO)₄ and (0.06 mol - 0.04 mol) 0.02 mol CO.

Now we can calculate the pressure of each gas after the reaction:

PNi(CO)₄ = n * R * T / V

PNi(CO)₄ = 0.01 mol * 0.082 (l amt / K mol) * 346K / 1.50 l = 0.189 atm

In the same way for CO:

P(CO) = 0.02 mol * 0.082 (l amt / K mol) * 346K / 1.50 l = 0.189 atm = 0.378 atm

The total pressure (Pt) in the bulb, according to Dalton´s law of partial pressures, is the sum of the pressures of each gas in the mixture:

Pt = PNi(CO)₄ + P(CO) = 0.189 atm + 0.378 atm = <u>0.567 atm.</u>

6 0

4 years ago

A temperature change in a reaction indicates_____.

natita [175]

A temperature change in a reaction indicates a chemical change

8 0

3 years ago

Read 2 more answers

Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib

Sergeeva-Olga [200]

Answer:

a) Heat of combustion of 1 g of methanol = -22.6 kJ = (-2.26 × 10) kJ

b) Heat of combustion of 1 g of ethanol = -29.7 kJ = (-2.97 × 10) kJ

c) Heat of combustion of 1 g of propanol = -33.5 kJ = (-3.35 × 10) kJ

Explanation:

a) The equation for the combustion of methanol is given as

CH₃OH + (3/2)O₂ → CO₂ + 2H₂O

The standard heat of combustion of methanol is given as -726 kJ/mol from literature.

But, 1 g of methanol will have the heat of combustion of the number of moles of methanol contained in 1 g of methanol.

Number of moles = (mass)/(molar mass)

Molar mass of (CH₃OH) = 32.04 g/mol

Number of moles = (1/32.04) = 0.03121 moles

1 mole of methanol has a heat of combustion of -726 kJ

0.03121 mole of methanol will have a heat of combustion of (0.03121 × -726) = -22.6 kJ

b) The equation for the combustion of ethanol is given as

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

The standard heat of combustion of ethanol is given as -1367.6 kJ/mol from literature.

But, 1 g of ethanol will have the heat of combustion of the number of moles of ethanol contained in 1 g of ethanol.

Number of moles = (mass)/(molar mass)

Molar mass of (C₂H₅OH) = 46.07 g/mol

Number of moles = (1/46.07) = 0.0217 moles

1 mole of ethanol has a heat of combustion of -1367.6 kJ

0.0217 mole of ethanol will have a heat of combustion of (0.03121 × -1367.6) = -29.7 kJ

c) The equation for the combustion of propanol is given as

C₃H₇OH + (9/2)O₂ → 3CO₂ + 4H₂O

The standard heat of combustion of propanol is given as -2020 kJ/mol from literature.

But, 1 g of propanol will have the heat of combustion of the number of moles of propanol contained in 1 g of propanol.

Number of moles = (mass)/(molar mass)

Molar mass of (C₃H₇OH) = 60.09 g/mol

Number of moles = (1/60.09) = 0.0166 moles

1 mole of propanol has a heat of combustion of -2020 kJ

0.0166 mole of propanol will have a heat of combustion of (0.0166 × -2020) = -33.5 kJ

Hope this Helps!!!

5 0

3 years ago

I need someone to help me with this. PLEASE, I'M BEGGING YOU

musickatia [10]

Answer:

D. 25%

Explanation:

8 0

3 years ago

I really need help with this ASAP​

I really need help with this ASAP