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Fynjy0 [20]
1 year ago
11

Beryllium has an atomic number of 4. How many neutrons does the isotope beryllium-9 have?.

Chemistry
1 answer:
VLD [36.1K]1 year ago
3 0

The number of neutrons possessed by an isotope beryllium-9 is 5.

<h3>What is neutron?</h3>

Neutron is a subatomic particle forming part of the nucleus of an atom and having no charge.

The number of neutrons in the atom of an element can be calculated by subtracting the number of protons or atomic number from the mass number.

That is; no. of neutrons = mass number - no of protons

According to this question, Beryllium has an atomic number or proton number of 4. The number of neutrons in an isotope of beryllium-9 is as follows:

no of neutrons = 9 - 4 = 5

Therefore, the number of neutrons possessed by an isotope beryllium-9 is 5.

Learn more about neutrons at: brainly.com/question/13131235

#SPJ1

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4 0
2 years ago
How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?
shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = \frac{no. of moles}{Volume (in L) of solution}

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = \frac{no. of moles}{Volume of solution in liter}

            0.0800 M = \frac{no. of moles}{0.05 L}

            no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = \frac{mass in grams}{molar mass}

             mass in grams = no. of moles \times molar mass of CuSO_{4}.5H_{2}O

                                       = 1.6 mol \times 249.68 g/mol

                                       = 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

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3 years ago
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AfilCa [17]
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3 years ago
how many covalent bonds does a phosphorus atom normally form, based on the number of valence electrons in the atom?
Alexandra [31]

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8 0
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