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lesya692 [45]
3 years ago
7

How much positive charge is in 1.4 kg of oxygen? The atomic weight (15.9994 g) of oxygen contains Avogadro’s number of atoms, wi

th each atom having 8 protons and 8 electrons. The elemental charge is 1.602 × 10−19 C and Avogadro’s number is 6.02
Chemistry
1 answer:
Cerrena [4.2K]3 years ago
5 0

Answer:

6.7511\times 10^7\ C

Explanation:

The atomic weight of oxygen = 15.9994 g

This mass corresponds to 1 mole of the oxygen atoms.

Thus,

15.9994 g mass of oxygen contains 6.02\times 10^{23} atoms of oxygen.

1.4 kg = 1400 g ( 1 kg = 1000 g)

So,

1400 g mass of oxygen contains \frac {6.02\times 10^{23}}{15.9994}\times 1400 atoms of oxygen.

Number of atoms in 1400 g of oxygen = 526.769754\times 10^{23}

Also, 1 atom of oxygen contains 8 protons

Charge of 1 proton = + 1.602\times 10^{-19}\ C

So, Charge on 1 atom of oxygen = 8\times 1.602\times 10^{-19}\ C

Thus,

Charge on 526.769754\times 10^{23} atoms of oxygen = 526.94476\times 10^{23}\times 8\times 1.602\times 10^{-19}\ C=6.7533\times 10^7\ C

Thus, positive charge in 1.4 kg of oxygen = 6.7511\times 10^7\ C

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daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)

The expression for enthalpy change is,

\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]

Putting the values we get :

\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]

-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]

H_f_{C_4H_{10}=-125kJ/mol

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0
3 years ago
Dinitrogen pentoxide is used in the preparation of explosives. If 7.93 mol of
Tpy6a [65]

The volume of O₂ produced: 84.6 L

<h3>Further explanation</h3>

Given

7.93 mol of  dinitrogen pentoxide

T = 48 + 273 = 321 K

P = 125 kPa = 1,23365 atm

Required

Volume of O₂

Solution

Decomposition reaction of dinitrogen pentoxide

2N₂O₅(g)→4NO₂(g)+O₂ (g)

From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :

= 0.5 x mol N₂O₅

= 0.5 x 7.93

= 3.965 moles

The volume of O₂ :

\tt V=\dfrac{nRT}{P}\\\\V=\dfrac{3.965\times 0.082\times 321}{1.23365}\\\\V=84.6~L

5 0
4 years ago
A 25.888 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 73.464 g of wa
brilliants [131]

Answer:

Weight % of NH₃ in the aqueous waste  = 2.001 %

Explanation:

The chemical equation for the reaction

\\\\NH_3} + HCl -----> NH_4Cl

Moles of HCl = Molarity × Volume

= 0.1039 × 31.89 mL × \frac{1 \  L}{1000 \ mL}

= 0.0033 mole

Total mass of original sample = 25.888 g + 73.464 g

= 99.352 g

Total HCl taken for assay = \frac{10.762 \ g}{99.352 \ g}

= 0.1083 g

Moles of NH₃ = \frac{0.0033 \  mol}{0.1083}

= 0.03047 moles

Mass of NH₃ = number of  moles × molar mass

Mass of NH₃ = 0.03047 moles × 17 g

Mass of NH₃  = 0.51799

Weight % of NH₃  = \frac{0.51799 \ g}{25.888 \ g} * 100%%

Weight % of NH₃ in the aqueous waste  = 2.001 %

4 0
3 years ago
Why are basic radicals classified into six groups?​
Alona [7]

Basic radicals (cations) have been divided into groups based on Ksp values

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