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3 years ago

What is the molarity of a nano3 solution if 25.0 ml of a 0.200 m nano3 solution is diluted?

1 answer:

6 0

samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness. In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

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The smell of dirty gym socks is caused by the compound caproic acid(contains H, C, O). Combustion of 0.844 g of caproic acid pro

adoni [48]

<u>Answer:</u> The molecular formula for the given organic compound is $C_2H_{32}O_4$

<u>Explanation:</u>

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:

Mass of $CO_2=0.784g$

Mass of $H_2O=1.92g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

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

So, in 0.784 g of carbon dioxide, $\frac{12}{44}\times 0.784=0.214g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

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

So, in 1.92 g of water, $\frac{2}{18}\times 1.92=0.213g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.844) - (0.214 + 0.213) = 0.417 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.214g}{12g/mole}=0.0134moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.213g}{1g/mole}=0.213moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.417g}{16g/mole}=0.0261moles$

<u>Step 2:</u> 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 0.0134 moles.

For Carbon = $\frac{0.0134}{0.0134}=1$

For Hydrogen = $\frac{0.213}{0.0134}=15.89\approx 16$

For Oxygen = $\frac{0.0261}{0.0134}=1.95\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 16 : 2

The empirical formula for the given compound is $CH_{16}O_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 116.2 g/mol

Mass of empirical formula = 60 g/mol

Putting values in above equation, we get:

$n=\frac{116.2g/mol}{60g/mol}=1.94\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(16\times 2)}O_{(2\times 2)}=C_2H_{32}O_4$

Hence, the molecular formula for the given organic compound is $C_2H_{32}O_4$

5 0

3 years ago

How many moles of chlorine gas and 120°C and 33.3 ATM would occupy a vessel of 12 L?

Mashutka [201]

<h3>Answer:</h3>

12.387 moles

<h3>Explanation:</h3>

We are given;

Temperature of chlorine, T = 120°C

But, K = °C + 273.15

Therefore, T = 393.15 K

Pressure, P = 33.3 Atm

Volume, V = 12 L

We are required to calculate the number of moles of chlorine gas,

To find the number of moles we are going to use the ideal gas equation;

PV = nRT

R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore, rearranging the formula;

n = PV÷RT

Hence;

n = (33.3 atm × 12 L) ÷ (0.082057 × 393.15 K)

= 12.387 moles

Therefore, the number of moles of chlorine are 12.387 moles

8 0

4 years ago

a bottle of acetone contains 55.0mL of acetone. If the acetone has a mass of 43.6 g. What is the density of acetone

Lady_Fox [76]

Answer:

792.72g/L

Explanation:

density formula

density = mass/volume in L

so the density of acetone = 43.6/0.055

792.72g/L

4 0

3 years ago

Read 2 more answers

Can anybody complete this for me?

NeTakaya

Answer:

I wish i could

Explanation:

3 0

3 years ago

Think about something very important to you that you would like to accomplish 10 years from now. Write a short-term, medium-term

Vinil7 [7]

Answer:

Goal: to propose a strategy to mitigate carcinogenic dyes present in wastewater.

Explanation:

Hello,

Short-term: Based on my interest about the treatment of carcinogenic effluents in the dyes industry, I'm planning to develop a research about how fast microorganisms could decompose such dyes in order to provide pure water-composed outlet streams. For now, I can suggest that they could work as long as they grow in a simulated dye-rich environment.

Medium-term: In such a way, the subsequent plan is to set up the reaction scheme and the reactor design stage for this bioprocess by running several test in order to collect the experimental data to back up the proposed mathematical model.

Long-term: Finally, the start-up of the whole process starting by the inlets and operation conditions, continuing by the innermost sub processes to decompose the dye and the subsequent adaptation of the outlet streams to produce drinkable water, will be carried out in order to help the environment and living things that are highly affected by those dyes.

Best regards.

5 0

3 years ago