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VARVARA [1.3K]
3 years ago
10

Which solvent water or carbon tetrachloride would you choose to dissolve each of the following?

Chemistry
1 answer:
ddd [48]3 years ago
4 0

Answer:

soorry i dont know this one.

Explanation:

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How do scientists test for a substance’s flammability in a substance
saveliy_v [14]

Answer:

Determining flammability can be as simple as holding a sample of the substance over a match. If it burns, it is flammable, leading to additional experiments to find more properties. Measuring the heat given off by the substance when it burns gives the heat of combustion.

Explanation:

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3 years ago
Name the following compound: CH3 – CH2 – CH2 – CH2 – CH2 – CH2 – CH2 – CH3 octyle octyne octane octene
ad-work [718]
8 carbons is "oct" prefix (like polygons)
No pi bonds is "ane" suffix.

put it together to get.... Octane

6 0
3 years ago
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The solubility of acetanilide in hot water (5.5 g/100 ml at 100∘C) is not very great, and its solubility in cold water (0.53 g/
Olin [163]

Answer:

89.4%

Explanation:

Initially, there is 5.0 of the acetanilide in 100 mL of water, then the solution is chilled at 0ºC. The solubility represents the amount that the solvent (water) can dissolve of the solute (acetanilide). So, at 0ºC, 100 mL of water can dissolve till 0.53 g of the compound, the rest will precipitate and will be recovered.

So, the mass that is recovered is 5.0 - 0.53 = 4.47 g

The percent recovery is:

(4.47/5)x100% = 89.4%

8 0
4 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>

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