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3 years ago

When can a compound be treated like an element when balancing chemical equations?

2 answers:

7 0

A compound can be treated like an element when balancing chemical equations when : A. when there is no change to the atoms in the compound

If they remain together during the reaction, you can treat a compound as a singe atom.

hope this helps

yuradex [85]3 years ago

4 0

Answer: Option (A) is the correct answer.

Explanation:

When two or more elements combine chemically to each other then it results in the formation of a substance known as a compound.

On the other hand, a substance that contains same atoms is known as an element.

For example, $F_{2}$ is a diatomic molecule and since it contains more than one atom so it is a compound also.

Therefore, we can conclude that when there is no change to the atoms in the compound then a compound can be treated like an element when balancing chemical equations.

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A beaker holds 962 g of a brine solution that is 6.20 percent salt. If 123g of water are evaporated from the beaker, how much sa

denpristay [2]

Answer:

13.687 grams of salt should be added

The total grams of 8.6% of brine solution produced is 852.687g

Explanation:

Solution mass= 962g

Salt= 6.2%

Water = 93.8%

962 gram of water is made up of:

902.356g ( due to vaporization which reduces mass)

= 902.356 - 123

= 779. 356g of water

59.644g of salt.

If we add x gram of salt for making brine solution up to 8.6%

=(59. 644g + x)g.of salt

% salt = Mass of Salt / Total mass of solution

= 0.086= 59.644 + x / 779.356 + 59.644 + x

= 59.644 + x / 839 + x

x= 13.687 g of salt

Grams of 8.6% brine solution will be:

Gram of water + total gram of salt added to form 8.6% brine solution.

= 779.356g +59.644g + 13.687g

= 852.687g

The total grams of 8.6% of brine solution produced is 852.687g

8 0

3 years ago

What trend does the first ionization energy follow going across the periodic table?

djyliett [7]

Explanation:

Ionization energy is defined as the energy necessary to remove an electron from a gaseous atom or ion.

Therefore, smaller is the size of an atom or ion more energy it needs to remove an electron because more is the charge on an ion smaller will be its size.

Hence, more will be the attraction between nucleus and valence electrons of the atom. So, more difficulty is faced by the atom to lose an electron. As a result, ionization energy will increase.

Across the period, there will be decrease in size of elements of the periodic table.

Thus, we can conclude that there will be increase in first ionization energy across the periodic table.

4 0

3 years ago

Read 2 more answers

Which process is NOT an exothermic process? A) Boiling B) Condensing C) Deposition D) Freezing

Elena-2011 [213]

The answer is D Freezing

5 0

3 years ago

What is the molarity of a solution in which 25.3 grams of potassium bromide is dissolved in 150. mL of solution?

irga5000 [103]

Answer:

~1.417M

Explanation:

Molarity=(number of moles of solute)/(litres of solution)

In this case, we need to find moles of potassium bromide.

Mass=25.3g

Molar mass= 119g/mol

moles=(mass/molar mass)

=(25.3)/(119)

=0.2126moles of potassium bromide

Molarity=(0.2126)/(150/1000)

~1.417M

Hope this helps:)

6 0

3 years ago

Read 2 more answers

Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....

Oduvanchick [21]

Answer:

Explanation:

$\mathbf{From \ the \ information \ given:} \\ \\ \mathbf{The \ rates \ of \ the \ f ollowing \ reactions \ can \ be \ expressed \ as \ follows:}$

(a)

$\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}$

(b)

$\mathbf{C_2H_6 \to C_2H_4 + H_2} \\ \\ \\ \mathbf{ -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}$

(c)

$\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} = \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt} }$

(d)

$\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\ \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}$

(e)

$\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\ \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}$

5 0

3 years ago