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Svetach [21]
3 years ago
13

How are the number of neutrons in the nucleus of an atom calculated?

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
7 0

Answer:

To calculate the number of neutrons in the nucleus of an atom is simple. You take the atomic, or proton number of the element, and you subtract it from the element's mass number.

Explanation:

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The symbol and group number of the period 5 transition element whos atoms have the fewest protons
worty [1.4K]

Answer:

Y and group number 17

Explanation:

Iodine is the chemical element of symbol I and atomic number 53, belonging to group 17 or of the halogens, in the periodic table. It is an element that we can find both in its molecular form, and in a diatomic way. In addition, iodine is a trace element and is used in the field of medicine and photography, as well as in some cases it is used as a dye. In the field of chemistry, iodine is the least reactive halogen element. Like the rest of the halogens, iodine also forms diatomic molecules, in this case the diiodine (I2).

Iodine is part of numerous compounds, despite being the element with the lowest reactivity in its group. It has some characteristics of metals, and their oxidation states are -1, + 1, + 3, + 5, + 7.

Despite its use, it is the halogen as a minor abundance, with only 0.14 ppm concentration in the earth's crust, however, in seawater its abundance is exactly 0.052 ppm.

8 0
4 years ago
A scientist claims to have a cooling apparatus kept at -100C by liquid nitrogen. Is this possible? Why or why not?
Vilka [71]

Answer:

It is not possible

Explanation:

Nitrogen is a light gas that find application in my areas including being used to keep things cold.It has a freezing point of -210°c so the scientist Nitrogen in the question at 100°c have already freezed and no longer a liquid.

5 0
4 years ago
Write the charge balance for an aqueous solution of arsenic acid, H3AsO4, in which the acid can dissociate to H2AsO24 , HAsO24 2
lilavasa [31]

Answer:

Kindly check the explanation section.

Explanation:

When the Arsenic acid,H3AsO4 is put in water it dissociates to form or give out 1 proton, 2 protons or 3 protons that is 1H^+, 2H^+ and 3H^+. Check the equation showing the dissociation reactions as given below;

H3AsO4 ===========> H^+ + AsO4^-

H3AsO4 ===========> 2H^+ + AsO4^2-

H3AsO4 ===========> 3H^+ + AsO4^3-

Therefore, the total charge balance is given as:

[H^+] =[OH^-] + [ AsO4^-] + 2[ AsO4^2-] + 3[ AsO4^3-].

Kindly note that the [OH^-] is from water.

8 0
3 years ago
A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
Snezhnost [94]

Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

                                                                 = 0.175 M

and,      moles of sodium propionic acid = (0.26 - 0.02) mol

                                                                  = 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

7 0
3 years ago
In which phases of matter are the atoms closely packed but still able to slide past east other
nignag [31]

✯Hello✯

Its a liquid :)

Have a nice day

7 0
3 years ago
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