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3 years ago

What is the atomic mass number of an isotope that has an atomic number of 5 and includes 6 neutrons?

Chemistry

2 answers:

irinina [24]3 years ago

8 0

The answer should be; 11

The atomic mass number is found by combining the number of protons and neutrons

Hope this helps :)

tatuchka [14]3 years ago

7 0

The correct answer is 11

Explanation:

In chemistry, each atom contains protons, neutrons, and electrons. The atomic mass number of an isotope (a variant of a chemical element) can be determined by adding the number of protons and neutrons. Additionally, the number of protons is equivalent to the atomic number that is usually the same according to the element. Additionally, the number of electrons has no or little influence on the atomic mass number. This implies, if an isotope has an atomic number of 5 it has 5 protons and this added to the 6 neutrons of the isotope makes the atomic mass number of this isotope 11 or 5 protons + 6 neutrons = 11 (atomic mass number).

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An ionic solid is placed in water. which information is described by the solubility product constant?

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How much energy is lost to condense 300. grams of steam at 100.C?

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<h3>Further explanation</h3>

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1. phase change(steam to water)

2. cool down(100 to 0 C)

1. phase change(condensation)

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3 years ago

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2 years ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

<u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

<u>For NaI:</u>

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

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3 years ago