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LUCKY_DIMON [66]
3 years ago
7

How many oxygen atoms are represented by the formula Fe(CIO4)3? iron(III) chlorate

Chemistry
2 answers:
Finger [1]3 years ago
7 0

Answer:

\boxed {\boxed {\sf C. \ 12 \ oxygen \ atoms }}

Explanation:

We are given the chemical formula:

Fe(ClO_4)_3

There are 3 elements here:

  • Fe: Iron
  • Cl: Chlorine
  • O: Oxygen

The question asks for the number of oxygen atoms, so we can just focus on the O in the formula.

The O has a subscript of 4, indicating there are 4 oxygen atoms in the compound. But the compound is also enclosed in parentheses with a subscript of 3. Therefore, there are 3 of the compounds with 4 oxygen atoms.

We can multiply 3 and 4.

  • 3*4= 12

There are <u>12 oxygen atoms.</u>

zysi [14]3 years ago
5 0

Answer:

12

Explanation:

the 4 by the element symbol O multiplied by the 3 on the outside of the parentheses

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A gas has a pressure of 5.7 atm at 100.0°C. What is its pressure at20.0°C (Assume volume is unchanged)
son4ous [18]

Answer:

\large \boxed{\text{4.5 atm}}

Explanation:

The volume and amount of gas are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ =5.7 atm; T₁ = 100.0 °C

p₂ = ?;          T₂ =  20.0 °C

Calculations:

1. Convert the temperatures to kelvins

T₁ = (100.0 + 273.15) K = 373.15

T₂ =  (20.0 + 273.15) K = 293.15

2. Calculate the new pressure

\begin{array}{rcl}\dfrac{5.7}{373.15} & = & \dfrac{p_{2}}{293.15}\\\\0.0153 & = & \dfrac{p_{2}}{293.15}\\\\0.0153\times 293.15 &=&p_{2}\\p_{2} & = & \textbf{4.5 atm}\end{array}\\\text{The new pressure will be $\large \boxed{\textbf{4.5 atm}}$}

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What is the speed of a car that traveled 40.2km in 2 hours?
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Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

3 0
3 years ago
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