Which indicator has to be used in order to get an emerald green color at a pH of 9?

What volume of water must be added to 11.1 mL of a pH 2.0 solution of HNO3 in order to change the pH to 4.0? A) 11.1 mL B) 89 mL

Varvara68 [4.7K]

Answer:

Volume of water that must be added is 1.10 L

Explanation:

pH measures the acidity or the alkalinity of a substance

It is given by;

pH = -log[H+]

Using this we can find the concentration of H+ ions in the acid

pH = 2 = -log[H+]

Therefore;

[H+] = 10^-2

= 0.01 M

But, since 1 mole HNO₃ ionizes to give 1 mole of H+, then the concentration of HNO₃ is equal to the concentration of H+ ([HNO₃] = [H+])

Therefore;

Initial [HNO₃] = 0.01 M

Initial volume of HNO₃ = 11.1 mL or 0.0111 L

We can then use dilution equation to find the final volume after dilution.

The final pH is 4

Therefore, [H+] = 10^-4

= 0.0001 M

Thus, the final concentration of HNO₃ is 0.0001 M

Using dilution equation;

M1V1 =M2V2

Thus; V2 = M1V1÷ M2

= (0.01 M× 0.0111 L)÷ 0.0001 M

= 1.11 L

This means the final total volume will 1.11 L or 1110 ml

Therefore; The volume of water added = 1110 ml - 11.1 ml

= 1098.9 ml or

= 1.0989 L

= 1.10 L(2 d.p.)

Hence, The volume of water that must be added is 1.10 L

5 0

2 years ago

Balance the following reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the coeffici

Sveta_85 [38]

2 NH3 + 5F2 = blank N2F4 + 6 HF

8 0

1 year ago

Read 2 more answers

Analyze: The first shell can hold a maximum of two electrons. How does this explain the valence of hydrogen

chubhunter [2.5K]

Answer:

See explanation

Explanation:

Hydrogen has a valency of +1 or -1. Its electronic configuration is 1s1.

The 1s sub-level (first shell) is known to hold two electrons. This means that hydrogen may either loose this one electron in the 1s level to yield H^+ or accept another electron into this 1s level to form H^- (the hydride ion).

The formation of the hydride ion completes the 1s orbital.

4 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

Kisachek [45]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

<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 $CO_2=39.01g$

Mass of $H_2O=10.65g$

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 39.01 g of carbon dioxide, $\frac{12}{44}\times 39.01=10.64g$ 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 10.65 g of water, $\frac{2}{18}\times 10.65=1.18g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.64 + 1.18) = 1.6 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 = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.64g}{12g/mole}=0.886moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.18g}{1g/mole}=1.18moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.6g}{16g/mole}=0.1moles$

<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 0.1 moles.

For Carbon = $\frac{0.886}{0.1}=8.86\approx 9$

For Hydrogen = $\frac{1.18}{0.1}=11.8\approx 12$

For Oxygen = $\frac{0.1}{0.1}=1.99\approx 2$

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

The ratio of C : H : O = 9 : 12 : 1

The empirical formula for the given compound is $C_9H_{12}O$

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 = 272.38 g/mol

Mass of empirical formula = 136 g/mol

Putting values in above equation, we get:

$n=\frac{272.38g/mol}{136g/mol}=2$

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

$C_{(2\times 9)}H_{(2\times 12)}O_{(2\times 1)}=C_{18}H_{24}O_2$

Hence, the empirical and molecular formula for the given organic compound is $C_9H_{12}O$ and $C_{18}H_{24}O_2$

6 0

2 years ago

What is the formal charge on the nitrogen in hydroxylamine, h2noh?

andrew11 [14]

<h3>Answer:</h3>

Formal Charge on Nitrogen is "Zero".

<h3>Explanation:</h3>

Formal Charge on an atom in molecules is calculated using following formula;

Formal Charge = [# of Valence e⁻s] - [e⁻s in lone pairs + 1/2 # of Bonding e⁻s]

As shown in attached picture of Hydroxylamine, Nitrogen atom is containing two electrons in one lone pair of electrons and six electrons in three single bonds with two hydrogen and one oxygen atom respectively.

Hence,

Formal Charge = [5] - [2 + 6/2]

Formal Charge = [5] - [2 + 3]

Formal Charge = 5 - 5

Formal Charge = 0 (zero)

Hence, the formal charge on nitrogen atom in hydroxylamine is zero.

5 0

1 year ago