Ask question

Ask question

All categories

3 years ago

15

A 10.0 mL sample of HNO3 was exactly neutralized by 13.5 mL of 1.0 M KOH. What is the molarity of the HNO3? Use the titrations f

ormula. Show all work.

1 answer:

Kruka [31]3 years ago

7 0

Answer: Thus molarity of $HNO_3$ is 1.35 M

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=1\\M_1=?M\\V_1=10.0mL\\n_2=1\\M_2=1.0M\\V_2=13.5mL$

Putting values in above equation, we get:

$1\times M_1\times 10.0=1\times 1.0\times 13.5\\\\M_1=1.35M$

Thus molarity of $HNO_3$ is 1.35 M

You might be interested in

What are 2 mixtures and explain your answer

liq [111]

Water and orange juice
Or orange juice and sodium chloride

3 0

3 years ago

Describe three different scenarios where a chemical process or reaction shows an increase in entropy.

Jet001 [13]

The request is characterized as knowing where things are and having the capacity to discover and utilize the things.
In a compound procedure, there is more issue, more entropy when the particles
1. warm up, increment in temperature. The atoms are more disorganized
2. get stirred up and must be isolated with exertion. Bedlam.
3. state changes, dissolves, vaporizes. The atoms are more turbulent
4. respond to frame a pack of various particles. More disorder

4 0

3 years ago

Write a balanced chemical equation (smallest integer coefficients possible) for the reaction between an acid and a base that lea

11Alexandr11 [23.1K]

Answer:

2HClO4(aq) + Ca(OH)2(aq) → Ca(ClO4)2(aq) + 2H2O(l)

Perchloric acid + Calcium hydroxide → Calcium perchlorate + Water.

Explanation:

This is a neutralization reaction where the acid, Perchloric acid reacts completely with an appropriate amount of base, aqueous Calcium hydroxide to produce salt, aqueous Calcium perchlorate and water, liquid H2O only.

During this reaction, the hydrogen ion, H+, from the HClO4 is neutralized by the hydroxide ion, OH-, from the Ca(OH)2 to form the water molecule, H2O.

Thus, it is called a neutralization reaction.

6 0

3 years ago

The azide ion, n−3, is a symmetrical ion, all of whose contributing structures have formal charges. draw three important contrib

stich3 [128]

Explanation:

Contributing structures are the resonating structures which are formed due to the delocalization of electrons in a molecule.

The azide ion that is $N^{-}_3$ , is a symmetrical ion, all of whose contributing structures have formal charges.

Lone pair of central nitrogen atom in azide ion is in conjugation with the neighboring nitrogen atoms.

Contributing structures of azide ion are drawn in the image attached.

5 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago