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Zigmanuir [339]
3 years ago
5

This one is actually 30 points chem quesitons

Chemistry
1 answer:
GuDViN [60]3 years ago
5 0

Answer:

The answers to your questions are below:

Explanation:

Question 31 Silicon and sulphur, because they are in the same period (3)

Question 34 the third option, density is not in the table

Question 37 group 1A alkali metals

Question 38 group 1A, oxygen group

Question 39 They are metalloids

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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
What happens to the energy of gas particles when an elastic collision takes place?
julia-pushkina [17]
The answer is D because there is no forces of attraction or repulsion between gas particles .
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Lake Eola is a sinkhole lake. What reasoning best describes how it formed?
lutik1710 [3]

Answer: A

Explanation: the first answer is more reasonable than the rest of them , hope this helped

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Read 2 more answers
Sub-groups on periodic table
marissa [1.9K]

Answer:

Subgroups S, P, D, F, including elements of blocks s, p, d, f, respectively.

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4 0
3 years ago
(LO 3N, 4G, 4O) Aluminum sulfate is also involved in dying fabrics. The gelatinous precipitate formed in the reaction with dilut
brilliants [131]

9.4 × 10⁻³ mg (0.73 mmoles) of Al(OH)₃ is formed

Explanation:

We have the following chemical reaction:

Al₂(SO₄)₃(aq) + 6 NaOH(aq) → 2 Al(OH)₃(s) + 3 Na₂SO₄(aq)

The precipitate mentioned by the problem is aluminium hydroxide Al(OH)₃.

Now to determine the number of moles of sodium hydroxide NaOH we use the following formula:

molar concentration =  number of moles / volume

number of moles = molar concentration × volume

number of moles of NaOH = 0.088 M × 25 mL = 2.2 mmoles

number of moles of Al₂(SO₄)₃ = 5.6 × 10⁻³ moles = 5.6 mmoles (found in the  problem text)

We see from the chemical reaction that 1 mole of Al₂(SO₄)₃ requires 6 moles of NaOH so 5.6 mmoles of Al₂(SO₄)₃ would require 6 times more NaOH which is 33.6 mmoles and we have only 2.2 mmoles. The limiting reactant will be NaOH.

Now we devise the following reasoning:

if        6 mmoles of NaOH produces 2 mmoles of Al(OH)₃

then  2.2 mmoles of NaOH produces X mmoles of Al(OH)₃

X = (2.2 × 2) / 6 = 0.73 mmoles of Al(OH)₃

mass of Al(OH)₃ = number of moles / molecular weight

mass of Al(OH)₃ = 0.73 / 78

mass of Al(OH)₃ =  9.4 × 10⁻³ mg

Learn more:

precipitation reaction

brainly.com/question/10400269

7 0
3 years ago
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