All categories

3 years ago

Which equation should you use to find the final concentration?

Chemistry

1 answer:

Anon25 [30]3 years ago

6 0

Answer:

Mf = MiVi / Vf

Explanation:

To obtain the answer to the question given, let us obtain the equation for the final concentration.

This can be obtained from the dilution equation as shown below :

MiVi = MfVf

Mi is the initial concentration

Mf is the final concentration

Vi is the initial volume

Vf is the final volume.

Making Mf the subject of the above equation, we can obtain the equation which can be used to calculate the final concentration. This is illustrated below:

MiVi = MfVf

Divide both side by Vf

Mf = MiVi / Vf

Therefore, the equation which can be used to calculate the final concentration is Mf = MiVi / Vf

You might be interested in

The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of

gavmur [86]

Answer: The empirical formula for the given compound is $C_3H_6O$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Conversion factor: 1 g = 1000 mg

Mass of $CO_2=6.32mg=0.00632g$

Mass of $H_2O=2.58g=0.00258g$

Mass of compound = 2.78 mg = 0.00278 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.00632 g of carbon dioxide, $\frac{12}{44}\times 0.00632=0.00172g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.00258 g of water, $\frac{2}{18}\times 0.00258=0.000286g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.00278) - (0.00172 + 0.000286) = 0.000774 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.00172g}{12g/mole}=1.43\times 10^{-4}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.000286g}{1g/mole}=2.86\times 10^{-4}moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.000774g}{16g/mole}=4.83\times 10^{-5}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $4.83\times 10^{-5}mol$

For Carbon = $\frac{1.43\times 10^{-4}}{4.83\times 10^{-5}}=2.96\approx 3$

For Hydrogen = $\frac{2.86\times 10^{-4}}{4.83\times 10^{-5}}=5.92\approx 6$

For Oxygen = $\frac{4.83\times 10^{-5}}{4.83\times 10^{-5}}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 6 : 1

Hence, the empirical formula for the given compound is $C_3H_{6}O_1=C_3H_6O$

3 0

3 years ago

A chemist conducted a study on ammonia NH3(g) into hydrogen gas and nitrogen gas. This reaction is represented by the following

Lelu [443]

Answer:

Explanation:

7 0

3 years ago

Speed is the rate of change of position expressed as _____ traveled per unit of time. A. direction B. meter C. displacement D. d

STatiana [176]

Answer:

Distance

Explanation:

7 0

3 years ago

Find mass of 3 moles of water

EleoNora [17]

Answer:

54 g

Explanation:

1 mole of water = H2O

mass of 1 mole of H2O= mass of h2 + mass of o

= 2× mass of h +mass of o

= 2×1+16 =18 g

1 mole of water = 18g

3moles of water = 18×3g= 54g

6 0

3 years ago

Given the chemical formulas of the following compounds, name each compound and state the rules you used to determine each name.

Svetlanka [38]

When naming an ionic compound, write the name of the cation, which is the metal first. Then, write the name of the anion, which is the nonmetal. However, you remove the last 2-3 letters and replace suffixes.

1. RbF --> Rubidium Fluoride
Change fluorine to fluoride
2. CuO --> Copper (II) Oxide
Change oxygen to oxide. Oxide has a charge of -2. Since no subscripts are written, it means they have the same opposite charge. So, we use Copper (II).
3. (NH₄)₂C₂O₄ ---> Ammonium Oxalate
NH₄ is ammonia, but we change it to ammonium for polyatomic ions.

5 0

3 years ago