Ask question

Ask question

All categories

2 years ago

15

A 10.00 mL sample of 8.50 M HNO3 solution is diluted to a new volume of 65.0 mL. What is the concentration of the dilute solutio

n

1 answer:

loris [4]2 years ago

4 0

Answer: 1.31M

Explanation:

V1 = 10mL

C1 = 8.5M

V2 = 65mL

C2 =?

C1V1 = C2V2

10 x 8.5 = C2 x 65

C2 = (10 x 8.5 ) /65

C2 = 1.31M

You might be interested in

A student wants to draw a model of an atom. Which statement describes how to find the number of neutrons to include in the model

san4es73 [151]

c is the correct answer

4 0

2 years ago

A group of students are wandering around a room. when their teacher claps, the students sit down wherever they are. this situati

Oksi-84 [34.3K]

an amorphous solid because the particles do not have a regular structure is the answer

8 0

2 years ago

Read 2 more answers

Match each set of quantum numbers to the correct subshell description by typing in the correct number. 1: n = 2, l = 0 2p: 2: n

Reil [10]

Answer:

4

2

1

5

3

Explanation:

4 0

2 years ago

Read 2 more answers

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

<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 $CO_2=9.06g$

Mass of $H_2O=5.58g$

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

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

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

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

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

<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 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

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

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

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

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 = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

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

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

Which of the following trends is indirectly proportional to effective nuclear charge, Zeff

Julli [10]

Answer : Option C) Atomic Size

Explanation : The atomic radius of the elements is found to be decreasing if we go from left to right in the modern periodic table. Accordingly, $Z_{eff}$ increases as the number of shielding electrons present in the atomic nucleus of the periodic elements which lies in the same row remains constant while the number of protons in each atomic shell increases.

The effective nuclear charge $Z_{eff}$ of an atom is defined as the net positive charge which is felt by the valence electron of the atomic element.

When $Z_{eff}$ is observed to decrease, it is seen that the atomic radius grows in size. So, it explains the inverse relationship between both. This phenomenon occurs, because there is more screening of the electrons from the nucleus taking place, which is observed due to decrease the attraction between the electron and the nucleus.

7 0

2 years ago

Read 2 more answers