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frutty [35]
3 years ago
5

How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602

× 10−19 C and Avogadro’s number is 6.023 × 1023 . Answer in units of C.
Chemistry
1 answer:
hichkok12 [17]3 years ago
3 0

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as 1.602 \times 10^{-19} C. Hence, we will calculate its number of moles as follows.

          Moles = \frac{mass}{\text{molar mass}}

                     = \frac{0.7 \times 1000 g}{7 g/mol}

                     = 100 mol

According to mole concept, there are 6.023 \times 10^{23} atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

        No. of atoms = 100 \times 6.023 \times 10^{23}

                               = 6.023 \times 10^{25} atoms

Since, it is given that charge on 1 atom is as follows.

                     3 \times 1.602 \times 10^{-19}C

                    = 4.806 \times 10^{-19}C

Therefore, charge present on 6.023 \times 10^{25} atoms will be calculated as follows.

    6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C

            28.95 \times 10^{6}C

Thus, we can conclude that a positive charge of 28.95 \times 10^{6}C is in 0.7 kg of lithium.

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3 0
2 years ago
PLEASE HELP!!!
trasher [3.6K]

Answer:

(a) oxygen

(b) 154g (to 3sf)

(c) 79.9% (to 3sf)

Explanation:

mass (g) = moles × Mr/Ar

note: eqn means chemical equation

(a)

moles of P = 84.1 ÷ 30.973 = 2.7152 moles

moles of O2 = 85÷2(16) = 2.65625 moles

Assuming all the moles of P is used up,

moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)

moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)

therefore there is insufficient moles of O2 and the limiting reactant is oxygen.

(b)

moles of P2O5 produced

= 2/5 (according to eqn) × 2.7152

= 1.08608moles

mass of P2O5 produced

= 1.08608 × [ 2(30.973) + 5(16) ]

= 154.164g

= approx. 154g to 3 sig. fig.

(c)

% yield = actual/theoretical yield × 100%

= 123/154 × 100%

= 79.870%

= approx. 79.9% (to 3sf)

4 0
2 years ago
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