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Akimi4 [234]
2 years ago
14

Adding one proton to the nucleus of an atom

Chemistry
1 answer:
Sonbull [250]2 years ago
7 0
Adding or removing protons from the nucleus changes the charge of the nucleus and changes that atom's atomic number. So, adding or removing protons from the nucleus changes what element that atom is! For example, adding a proton to the nucleus of an atom of hydrogen creates an atom of helium.
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Why is it important to ensure that treated water remains safe to drink when it is stored after treatment? what is one way to mak
yarga [219]


It is important to ensure that treated water remains safe to drink because water does not last forever as it can gain bacteria and organisms in it. To make sure storage of water is safe is to simply add chlorine again over a period of time.

-never store in direct sunlight

-containment of the water is clean

-make sure chemicals or anything that can contaminate it doesn't come near it

7 0
2 years ago
You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number
irina1246 [14]

Answer: 9.11\times 10^{-31}kg

Explanation:

Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.

All non-zero numbers are always significant.

All zero’s between integers are always significant.

All zero’s after the decimal point are always significant.

All zero’s preceding the first integers are never significant.

Thus 9.11\times 10^{-31}kg has three significant figures

7 0
2 years ago
A copper rod that has a mass of 200.0 g has an initial temperature of 20.0°C and is heated to 40.0°C. If 1,540 J of heat are nee
Romashka [77]

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)</span>

1540 = 200.0 (C)(40 - 20)

<span> <span>C = 0.385 J / g C</span></span>

<span><span>
</span></span>

<span><span>Hope this answers the question. Have a nice day.</span></span>

8 0
2 years ago
Read 2 more answers
1. After clamping a buret to a ring stand, you notice that the set-up is tippy and unstable. What should you do to stabilize the
vampirchik [111]

Answer:

A. Move the buret clamp to a ring stand with a larger base.

Explanation:

The ring stands are used to hold burettes, light in weight to avoid loss of stability, that is why it is necessary to change the size of the ring stand so that it can support the buret that we are going to use.  It is not recommended to balance it with the hand since it would give us an inaccurate result in the titration.

7 0
2 years ago
When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing po
SOVA2 [1]

The given question is incomplete. The complete question is as follows.

When 70.4 g of benzamide (C_{7}H_{7}NO) are dissolved in 850 g of a certain mystery liquid X, the freezing point of the solution is 2.7^{o}C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH_{4}Cl) are dissolved in the same mass of X, the freezing point of the solution is 9.9^{o}C lower than the freezing point of pure X.

Calculate the Van't Hoff factor for ammonium chloride in X.

Explanation:

First, we will calculate the moles of benzamide as follows.

    Moles of benzamide = \frac{mass}{\text{Molar mass of benzamide}}

                    = \frac{70.4 g}{121.14 g/mol}

                    = 0.58 mol

Now, we will calculate the molality as follows.

     Molality = \frac{\text{moles of solute (benzamide)}}{\text{solvent mass in kg}}

                   = \frac{0.58 mol}{0.85 kg}

                   = 0.6837

It is known that relation between change in temperature, Van't Hoff factor and molality is as follows.

      dT = i \times K_{f} \times m,

where,      dT = change in freezing point = 2.7^{o}C

                  i = van't Hoff factor = 1 for non dissociable solutes

      K_{f} = freezing point constant of solvent

                m = 0.6837

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

             dT = i \times K_{f} \times m,

            2.7^{o}C = 1 \times K_{f} \times 0.6837 m

            K_{f} = 3.949 C/m

Now, we use this K_{f} value for calculating i for NH_{4}Cl

So, moles of ammonium chloride are calculated as follows.

 Moles of NH_{4}Cl = \frac{70.4 g}{53.491 g/mol}

                            = 1.316 mol

Hence, calculate the molality as follows.

    Molality = \frac{1.316 mol}{0.85 kg}

                  = 1.5484

It is given that value of change in temperature (dT) = 9.9^{o}C. Thus, calculate the value of Van't Hoff factor as follows.

              dT = i \times K_{f} \times m

   9.9^{o}C = i \times 3.949 C/m \times 1.5484 m

                     i = 1.62

Thus, we can conclude that the value of van't Hoff factor for ammonium chloride is 1.62.

5 0
2 years ago
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