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vivado [14]
2 years ago
13

Boron occurs naturally as two isotopes. What is the difference between these isotopes?

Chemistry
1 answer:
Murrr4er [49]2 years ago
8 0

Answer:

Number of neutrons and stability

Explanation:

An isotope of an element is basically the same element but with different number of neutrons. For example here, boron can exist in the forms of boron-10 and boron-11, and so the latter would have one more neutron than the former one.

Adding an extra neutron may or may not disrupt the strong force that much, and so the half-life and stability of the new isotope can be slightly different than its most stable one.

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What happens to an object at rest if balanced forces act upon it?
OLga [1]

Newton's first law of motion states that an object at rest will remain at rest unless an unbalanced force acts on it. If you apply balanced forces on the object there would be no net force. The body does not accelerate but instead stays at rest.

Another way to look at this problem is to use Newton's second law of motion. The first law states that F = m\times a, where a is the accelerationF is the net force and m is the mass of the object.

When F is zero, the acceleration of the object is zero. This means that if the object had a velocity of zero before the balanced forces started acting, the velocity will stay at zero after the balanced forces begin to act. If  the object was moving at a constant velocity before the balanced forces started acting on it, it would continue at that constant velocity after the balanced forces begin to act.

4 0
3 years ago
How many neutrons are in 7.00 g of 13c?
Minchanka [31]
You multiply the number of atoms by 12 to get how many electrons (since each atom has 12 electrons in it)
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3 years ago
What is the molar mass of C3N4? (Do NOT round this number.)
vichka [17]

Answer:

92.06 g/mol

Explanation:

4 0
2 years ago
How much of the original amount of an isotope is present after a period of four half lives
vfiekz [6]

After 2 half-lives there will be 25% (1/4th) of the original isotope, and 75% (3/4 th) of the decay product

<h3>What is Half life period ?</h3>

A half life is a measurement of the slope of an exponential decay function.

It is also defined as, the time it takes to halve the concentration of something in a process.

Each half life you will have half of what you had at the beginning of a given half life.

Learn more about half life here ;

brainly.com/question/9654500

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7 0
1 year ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
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