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3 years ago

Radioactive isotopes of an atom are

Chemistry

2 answers:

nikdorinn [45]3 years ago

5 0

Less stable should be the answer.as I think
I hope this helps you

Virty [35]3 years ago

4 0

Answer:b/ less stable

Explanation:

Many elements that have one or more isotopes which are radioactive in nature have the nucleus which remains in an unstable state. The unstable nucleus breaks down, decay and emit the radiation. The nucleus in the atoms of these elements are unstable because it consists of unstable ratio of protons to neutron.

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Vilka [71]

Answer:

what does this mean lol

Explanation:

6 0

3 years ago

100 points !!!! Write your answer beside each number. You should have 13 items labeled in all.

Arte-miy333 [17]

Answer:

Explanation:

1) Valence electrons

2) Electrons

3) 2 electrons on K-shell

4) 8 electrons on L-Shell

5) 18 electrons on M-shell

6) nucleus

7) Proton

8) neutron

9) principal energy level

10) atomic number

11) symbol of the atom

12) name of the atom

13) atomic mass

5 0

3 years ago

5. A box with a volume of 22.4 L contains 1.0 mol of nitrogen and 2.0 mol of hydrogen at 0°C. Which of the following statements

love history [14]

B. The partial pressure of N2 is 101 kPa

<h3>Further explanation</h3>

Given

volume = 22.4 L

1.0 mol of nitrogen and 2.0 mol of hydrogen at 0°C

Required

Total pressure and partial pressure

Solution

Ideal gas law :

PV = nRT

n total = 3 mol

T = O °C + 273 = 273 K

P = nRT/V

P = 3 x 0.08205 x 273 / 22.4

P total = 3 atm = 303,975 kPa

P Nitrogen = 1/3 x 303.975 = 101.325 kPa

P Hydrogen = 2/3 x 303.975 = 202.65 kPa

7 0

3 years ago

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit

natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.

7 0

3 years ago

At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepar

Rudiy27

Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

moles methanol = 4.40 moles

Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

Total moles = 2.115 moles + 4.40 moles

Total moles = 6.515 moles

Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

P(acetic acid) = 0.325 * 226

P(acetic acid) = 73.45 torr ≈73.5

We can control this by calculating the partial pressure of methanol

mole fraction of methanol = (6.515-2.115)/6.515 = 0.675

P(methanol) = 0.675 * 226 = 152.55

226 - 152.55 = 73.45 torr

The partial pressure of acetic acid is 73.5 torr

4 0

3 years ago