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4 years ago

Which statement best describes balancing equations and the law of conservation of mass?

Chemistry

2 answers:

stellarik [79]4 years ago

8 0

Answer:

The number of atoms is the same in the reactants and in the products, and the total mass is the same in the reactants and

in the products- second choice

Licemer1 [7]4 years ago

4 0

Answer:

The correct answer is: <u><em>"The number of atoms is the same in the reactants and in the products, and the total mass is the same in the reactants and in the products"</em></u>

Explanation:

The Law of Conservation of Matter is also called the law of conservation of mass or the Law of Lomonósov-Lavoisier. This law postulates that "the mass is not created or destroyed, only transformed." This means that the reagents interact with each other and form new products with physical and chemical properties different from those of the reagents because the atoms of the substances are ordered differently. But the amount of matter or mass before and after a transformation (chemical reaction) is always the same, that is, the quantities of the masses involved in a given reaction must be constant at all times, not changing in their proportions when the reaction ends. In other words, then the mass before the chemical reaction is equal to the mass after the reaction. The exception to the rule is nuclear reactions, in which it is possible to convert mass into energy and vice versa.

An example of this law is the combustion of hydrocarbons, in which the fuel can be seen burning and "disappearing", when in truth it will have been transformed into invisible gases and released energy.

This law is what allows the equations to be balanced or chemical reaction, to maintain the conservation of mass.

So, <u><em>the correct answer is "The number of atoms is the same in the reactants and in the products, and the total mass is the same in the reactants and in the products"</em></u>

You might be interested in

In the reaction ___h2+n2 ->2 NH3, what coefficient should be placed in front of H2 to balance the reaction?

Nataly_w [17]

In order to balance this equation you need to count each element and how many of the individual elements are in the equation.
_H2+N2=2 NH3
You multiply the 2 (Which is the coefficient) by the 3 (which is the subscript) This would equal 6 which indicated there are 6 hydrogen atoms on the right side so the left side should also have 6 hydrogen atoms

The missing coefficient on the left side must multiple the 2 to become 6 hydrogen
Answer=3

7 0

3 years ago

Read 2 more answers

The one factor that a scientist changes in an experiment is called the

Gala2k [10]

Answer:

It might be responding variable.

8 0

3 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

4 years ago

Any chemical or physical change that absorbs energy 1. cannot occur spontaneously. 2. forms products with less mass than the ini

grigory [225]

Answer:

3. is an endothermic process.

Explanation:

Generally, reactions can be classified into two. Depending on whether they absorb heat -energy or release energy.

Endothermic Reactions are those reactions in which the reaction absorb heat form the surroundings.

Exothermic reactions are those reactions that release heat to the surroundings.

This means the correct option is;

3. is an endothermic process.

8 0

3 years ago

For the following reaction, 42.2 grams of potassium hydrogen sulfate are allowed to react with 21.4 grams of potassium hydroxide

ASHA 777 [7]

Answer:

53.99g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

KHSO4(aq) + KOH(aq) —> K2SO4(aq) + H2O(l)

Step 2:

Determination of the masses of KHSO4 and KOH that reacted and the mass of K2SO4 produced from the balanced equation.

This is illustrated below:

Molar mass of KHSO4 = 39 + 1 + 32 + (16x4) = 136g/mol

Mass of KHSO4 from the balanced equation = 1 x 136 = 136g

Molar mass of KOH = 39 + 16 + 1 = 56g/mol

Mass of KOH from the balanced equation = 1 x 56 = 56g

Molar mass of K2SO4 = (39x2) + 32 + (16x4) = 174g/mol

Mass of K2SO4 from the balanced equation = 1 x 174 = 174g.

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH to produce 174g of K2SO4

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH.

Therefore, 42.2g of KHSO4 will react with = (42.2 x 56)/136 = 17.38g of KOH.

From the above calculations, we can see that only 17.38g out of 21.4g of KOH given was needed to react completely with 42.2g of KHSO4.

Therefore, KHSO4 is the limiting reactant and KOH is the excess reactant.

Step 4:

Determination of the maximum mass of K2SO4 produced from the reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction. The limiting reactant is KHSO4 and the maximum amount of K2SO4 produced can be obtained as follow:

From the balanced equation above, 136g of KHSO4 reacted to produce 174g of K2SO4.

Therefore, 42.2g of KHSO4 will react to produce = (42.2 x 174)/136 = 53.99g of K2SO4.

Therefore, the maximum amount of K2SO4 produced is 53.99g.

8 0

3 years ago