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4 years ago

How many atoms are in the product of the following reaction? 2KClO3 ⇋ 2KCl + 3O2 6 8 5 10

Chemistry

1 answer:

denis-greek [22]4 years ago

6 0

<u>Answer:</u>

<em>10 is the answer.</em>

<u>Explanation:</u>

Balancing is making the number of atoms of each element same on both the sides (reactant and product side).

To find the number of atoms of each element we multiply coefficient × subscript

For example $5 Ca_1 Cl_2$ contains

5 × 1 = 5 , Ca atoms and

5 × 2 = 10, Cl atoms

If there is a bracket in the chemical formula

For example $3Ca_3 (P_1 O_4 )_2$ we multiply coefficient × subscript × number outside the bracket.......... to find the number of atoms (Please note: 3 is the coefficient, and if there is no number given then 1 will be the coefficient )

So 3 × 3 = 9, Ca atoms

3 × 1 × 2 = 6, P atoms

3 × 4 × 2 = 24, O atoms are present.

Product side contains $2KCl+3O_2$

2, K

2, Cl

6, O

So totally 2 + 2 + 6 = 10 atoms present on the product side

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The average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

<h3>What is Atomic mass? </h3>

Atomic mass is defined as the whole mass of an atom.

It is also defined as the sum of atomic number and number of neutrons.

Atomic mass = Atomic number + neutrons

<h3>What is Isotopes?</h3>

Isotopes are the those element which have same atomic number but have different mass number and number of neutrons.

The average relative atomic mass can be calculated as

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2 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

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$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$

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