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Valentin [98]
3 years ago
8

Elements in the same column of the periodic table share:

Chemistry
1 answer:
avanturin [10]3 years ago
7 0
B. the same number of protons.
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The substance that can not be broken down by chemical means is antimony.
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A mains radio turns ________ energy into ______ energy and ______ energy.
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2 years ago
Suppose the formation of nitryl fluoride proceeds by the following mechanism: step elementary reaction rate constant (g) (g) (g)
ziro4ka [17]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a

  2NO_2 _{(g)} +  F_2_{(g)} ---->  2NO_2 F_{(g)}

b

  r = \frac{ k [NO_2]^2 [F_2]}{[NO_2F]}

Explanation:

      From the question we are told that

          The formation mechanism is  

                      NO_2_{(g)} + F_2 _{(g)} ----> NO_2 F_{(g)} + F_{(g)}

                      F_{(g)} +  NO_2 _{(g)} ---> NO_2 F_{(g)}

The overall balanced equation is

              2NO_2 _{(g)} +  F_2_{(g)} ---->  2NO_2 F_{(g)}

We combined the first reactant and the last product and the balanced the number of mole

     The observable rate law is

                 r = \frac{ k [NO_2]^2 [F_2]}{[NO_2F]}

This rate law is derived from the balanced chemical equation

     

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3 years ago
1.) Why does the heart rate speed up during exercise?
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At a wastewater treatment plant, FeCl3(s) is added to remove excess phosphate from the effluent. Assume the following reactions
RoseWind [281]

Answer : The concentration of Fe^{3+} needed is, 2.37\times 10^4M

Explanation :

First we have to calculate the mole of phosphate.

As we are given that, 1 mg P/L that means, 1 mg of phosphate present in 1 L of solution.

\text{Moles of phosphate}=\frac{\text{Mass of phosphate}}{\text{Molar mass of phosphate}}

Molar mass of phosphate = 94.97 g/mole

\text{Moles of phosphate}=\frac{1mg}{94.97g/mol}=\frac{0.001g}{94.97g/mol}=1.053\times 10^{-5}mol

Now we have to calculate the concentration of phosphate.

\text{Concentration of phosphate}=\frac{\text{Moles of phosphate}}{\text{Volume of solution}}

\text{Concentration of phosphate}=\frac{1.053\times 10^{-5}mol}{1L}=1.053\times 10^{-5}mol/L

Now we have to calculate the concentration of Fe^{3+}.

The second equilibrium reaction is,

FePO_4\rightleftharpoons Fe^{3+}+PO_4^{3-}

The solubility constant expression for this reaction is:

K_{sp}=[Fe^{3+}][PO_4^{3-}]

Given: K_{sp}=\frac{1}{4}

\frac{1}{4}=[Fe^{3+}]\times 1.053\times 10^{-5}mol/L

[Fe^{3+}]=2.37\times 10^4M

Thus, the concentration of Fe^{3+} needed is, 2.37\times 10^4M

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4 years ago
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