Which of the following is not an important use of geological hazard maps?

A 34.53 ml sample of a solution of sulfuric acid, h2s04, reacts with 27.86 ml of 0.08964 m naoh solution. calculate the molarity

Gnoma [55]

The balanced equation between NaOH and H₂SO₄ is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of moles of NaOH moles reacted = molarity of NaOH x volume
number of NaOH moles = 0.08964 mol/L x 27.86 x 10⁻³ L = 2.497 x 10⁻³ mol
according to molar ratio of 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
therefore 2.497 x 10⁻³ mol of NaOH reacts with - 1/2 x 2.497 x 10⁻³ mol of H₂SO₄
number of moles of H₂SO₄ reacted - 1.249 x 10⁻³ mol
Number of H₂SO₄ moles in 34.53 mL - 1.249 x 10⁻³ mol
number of H₂SO₄ moles in 1000 mL - 1.249 x 10⁻³ mol / 34.53 x 10⁻³ L = 0.03617 mol
molarity of H₂SO₄ is 0.03617 M

6 0

3 years ago

20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator

Eduardwww [97]

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to 5.6

Explanation:

The equation for the reaction is :

$C_2H_5NH_2_(_a_q_) + H^+_(_a_q_) --- C_2H_5NH_{3(aq)}^+$

concentration of $C_2H_5NH_{2(aq)$ = 10%

10 g of $C_2H_5NH_{2(aq)$ in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of $C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL$

= 2.22 M

number of moles of $C_2H_5NH_{2(aq)$ in 20 mL can be determined as:

$= 20 mL \times 2.22 M= 44*10^{-3} mole$

Concentration of $C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}$

= 2.22 M

Similarly, The pKa Value of $C_2H_5NH_{2(aq)$ is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

$pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]$

$pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]$

$pH = 5.21$

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

8 0

3 years ago

Which of the following are considered pure substances?

Helen [10]

B.Elements

Explanation: they cannot be separated

8 0

3 years ago

A gas mixture contains oxygen, nitrogen, and carbon dioxide. It has a pressure of 250 mmHg. If the pressure of oxygen is 50 mmHg

kondor19780726 [428]

Answer:

$P_{CO_2}=25mmHg$

Explanation:

Hello there!

In this case, according to the Dalton's law, which explains that the total pressure of a gaseous system equals the sum of the partial pressures of the gases composing, for the gaseous mixture composed by oxygen, nitrogen and carbon dioxide it would be possible to write:

$P_{tot}=P_{N_2}+P_{O_2}+P_{CO_2}$

Now, given the pressure of the system and those of oxygen and nitrogen, we calculate that of carbon dioxide as shown below:

$P_{CO_2}=P_{tot}-P_{N_2}-P_{O_2}\\\\P_{CO_2}=250mmHg-50mmHg-175mmHg\\\\P_{CO_2}=25mmHg$

Best regards!

4 0

2 years ago

The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 10−8 L/mol/s. What is the instantaneou

pochemuha

Answer: Rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For a reaction : $A\rightarrow products$

$Rate=k[A]^x$

k= rate constant

x = order of the reaction = 2

$Rate=4.71\times 10^{-8}L/mol/s\times {5.55\times 10^{-4}M}^2$

$Rate=1.45\times 10^{-14}$

Thus rate of decomposition of acetaldehyde in a solution is $1.45\times 10^{-14}mol/Ls$

6 0

3 years ago