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Schach [20]
3 years ago
14

When the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coeffici

ents are: ___ bromine trifluoride (g) ___ bromine (g) + ___ fluorine (g)
Chemistry
1 answer:
Sergio [31]3 years ago
7 0

<u>Answer:</u> The molecular balanced equation is written below

<u>Explanation:</u>

A molecular equation is defined as the chemical equation in which the ionic compounds are written as molecules rather than component ions.

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

The balanced chemical equation for the decomposition of bromine trifluoride follows:

2BrF_3\rightarrow Br_2+3F_2

By Stoichiometry of the reaction:

2 moles of bromine trifluoride produces 1 mole of bromine gas and 3 moles of fluorine gas

Hence, the molecular balanced equation is written above.

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Find the molar mass of CaCO3 then subtract the molar mass what it originally weighed and the loss of mass. Hopefully this works!
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Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5​
lubasha [3.4K]

The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

K_{a}=\frac{[H^{+}][HB] }{[reactant]}

[HB] is the concentration of base.

8 * 10^{-5} =\frac{x^{2}  }{0.1}\\\\\\x^{2} = 8 * 10^{-5}*0.1

x^{2} = 0.08 * 10^{-4}\\ \\x = 0.283*10^{-2}

Then

pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54

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Iodide can react to Pb²⁺ to make a solid compound.

4 0
3 years ago
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