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2 years ago

Grams in 1.083e+24 Molecules Na BR Please help explain why

Chemistry

1 answer:

adoni [48]2 years ago

6 0

Answer:

185 gms o NaBr

Explanation:

Na Br mole weight = 22.989 +79.904 = <u>102.893 gm/mole</u>

1.083 x 10^24 molecules / 6.022 x 10^23 molecules / mole = <u>1.798 moles</u>

102.893 * 1.798 = 185 gms

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Sort these species into isoelectronic groups. It doesn\'t matter which group goes in which box, so long as the correct species a

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Answer : The isoelectronic groups are:

$He,Be^{2+},Li^{+}$

$N^{3-},Mg^{2+}$

$S^{2-},Ti^{4+},K^{+}$

Explanation :

Isoelectronic : It is defined as the compound or molecule having the same number of electrons and the same number of electronic structure.

The element is helium. The number of electrons are 2.
The element is beryllium. The number of electrons are 4. The number of electrons in $Be^{2+}$ = 4 - 2 = 2
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The element is nitrogen. The number of electrons are 7. The number of electrons in $N^{3-}$ = 7 + 3 = 10
The element is neon. The number of electrons are 10.
The element is sulfur. The number of electrons are 16. The number of electrons in $S^{2-}$ = 16 + 2 = 18
The element is magnesium. The number of electrons are 12. The number of electrons in $Mg^{2+}$ = 12 - 2 = 10
The element is titanium. The number of electrons are 22. The number of electrons in $Ti^{4+}$ = 22 - 4 = 18
The element is potassium. The number of electrons are 19. The number of electrons in $K^{+}$ = 19 - 1 = 18

The isoelectronic groups are:

$He,Be^{2+},Li^{+}$

$N^{3-},Mg^{2+}$

$S^{2-},Ti^{4+},K^{+}$

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The enthalpy of fusion of solid n-butane is 4.66 kJ/mol. Calculate the energy required to melt 58.3 g of solid n-butane.

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Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

Explanation :

First we have to calculate the moles of n-butane.

$\text{Moles of n-butane}=\frac{\text{Mass of n-butane}}{\text{Molar mass of n-butane}}$

Given:

Molar mass of n-butane = 58.12 g/mole

Mass of n-butane = 58.3 g

Now put all the given values in the above expression, we get:

$\text{Moles of n-butane}=\frac{58.3g}{58.12g/mol}=1.00mol$

Now we have to calculate the energy required.

$Q=\frac{\Delta H}{n}$

where,

Q = energy required

$\Delta H$ = enthalpy of fusion of solid n-butane = 4.66 kJ/mol

n = moles = 1.00 mol

Now put all the given values in the above expression, we get:

$Q=\frac{4.66kJ/mol}{1.00mol}=4.66kJ$

Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ

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