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3 years ago

What is the percent of nitrogen in dinitrogen pentoxide?

Chemistry

1 answer:

Lostsunrise [7]3 years ago

4 0

Answer: D.) 25.9%

Explanation:

Dinitrogen pentoxide chemical formular : N2O5

Calculating the molar mass of N2O5

Atomic mass of nitrogen(N) = 14

Atomic mass of oxygen(O) = 16

Therefore molar mass :

N2O5 = (2 × 14) + (5 × 16) = 28 + 80 = 108g/mol

Percentage amount of elements in N205:

NITROGEN (N) :

(Mass of nitrogen / molar mass of N2O5) × 100%

MASS OF NITROGEN = (N2) = 2 × 14 = 28

PERCENT OF NITROGEN : (28/108) × 100%

0.259259 × 100%

= 25.925%

= 25.9%

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A scientist measures the standard enthalpy change for the following reaction to be 67.9 kJ:

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Answer: The standard enthalpy of formation of $H_2O(g)$ is -252.1 kJ/mol.

Explanation:

The balanced chemical reaction is,

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]$

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$\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]$

$67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]$

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3 years ago

Define electronegativity. Where on the periodic table is the highest? Explain why this is the case.

geniusboy [140]

Answer:

Explanation:

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Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

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$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

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

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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For the chemical reaction , I identify the reactant and the products .

svetlana [45]

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