How many atoms of potassium and how

Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

3 years ago

You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar

djverab [1.8K]

1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

<em>A</em> = <em>ε</em> $\cdot l \cdot c$ by the Beer-Lambert law, where

<em>A</em> the absorbance,
$l$ the path length,
<em>ε</em> the molar absorptivity of the solute, and
$c$ concentration of the solution.

<em>A</em> and <em>ε </em>are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

6 0

3 years ago

The greater the mass of an object

andreyandreev [35.5K]

The greater the mass of an object, the greater<span> its gravitational force.
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Plz chose as brainliest answer plz :) <#

6 0

3 years ago

How many grams of lithium chloride can be produced from 14.3 g of lithium chlorate when it decomposes into lithium chloride and

vampirchik [111]

6.7 grams of lithium chloride will be produced.

<h3><u>Explanation:</u></h3>

Lithium chlorate is LiClO₃ and lithium chloride is LiCl. The reaction is,

2LiClO₃ = 2LiCl +3O₂.

So here, 2 moles of lithium chlorate produces 2 moles of lithium chloride.

Or, one molecule of lithium chlorate will produce one mole of lithium chloride.

Molecular weight of lithium chlorate = $6.9 + 35.5 + 16\times3$ = 90.4.

So, 14.3 grams of lithium chlorate has 0.16 moles of lithium chlorate.

Thereby, moles of lithium chloride produced is 0.16 moles.

Molecular weight of lithium chloride = $6.9+35.5$ = 42.4 grams.

So weight of lithium chloride produced = $42.4 \times 0.16$ = 6.7 grams.

Thus, weight of lithium chloride produced will be 6.7 grams.

3 0

3 years ago

Give the nuclear symbol for the isotope of beryllium for which a=10? enter the nuclear symbol for the isotope (e.g., 42he).

galben [10]

An isotope is defined as an element that has the same number of protons as the common element but different number of neutrons. In this case, a beryllium atom has a molar weight of 10 amu. Thus, there are 4 protons and 6 neutrons. The nuclear symbol of Be-10 is 4 Be10

8 0

3 years ago