All categories

3 years ago

- How many moles of HCl will be needed to neutralize 200.0 ml of a 2.5 M NaOH solution?

Chemistry

1 answer:

dalvyx [7]3 years ago

4 0

Answer:

0.5 mol HCl

Explanation:

2.5 mol/L *0.200 L = 0.5 mol NaOH

NaOH + HCl ---> NaCl + H2O

from reaction 1 mol 1 mol

given 0.5mol x mol

x= 0.5*1/1= 0.5 mol HCl

You might be interested in

If a car travels 800m in 30 seconds how fast is it going? HELP MEEE!!

Paul [167]

Answer:

Explanation:

The speed of the car can be found by using the formula

$v = \frac{d}{t} \\$

d is the distance

t is the time taken

From the question we have

$v = \frac{800}{30} = \frac{80}{3} \\ = 26.66666...$

We have the final answer as

Hope this helps you

7 0

3 years ago

Suppose a current of is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for s

WARRIOR [948]

The given question is incomplete. The complete question is:

Suppose a current of 0.920 A is passed through an electroplating cell with an aqueous solution of agno3 in the cathode compartment for 47.0 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell.

Answer: 0.0484 g

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 0.920 A

t= time in seconds = 47.0 sec

$Q=0.920A\times 47.0s=43.24C$

$AgNO_3\rightarrow Ag^++NO_3^-$

$Ag^++e^-\rightarrow Ag$

96500 Coloumb of electricity electrolyzes 1 mole of Ag

43.24 C of electricity deposits = $\frac{1}{96500}\times 43.24=0.00045moles$ of Ag

$\text{ mass of Ag}={\text{no of moles}\times {\text{Molar mass}}=0.00045mol\times 108g=0.0484g$

Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0484 g

6 0

3 years ago

Caffeine (C8H10N4O2) is a weak base with a pKb of 10.4. Calculate the pH of a solution containing a caffeine concentration of 41

RSB [31]

Answer:

pH → 7.47

Explanation:

Caffeine is a sort of amine, which is a weak base. Then, this pH should be higher than 7.

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

1 mol of caffeine in water can give hydroxides and protonated caffeine.

We convert the concentration from mg/L to M

415 mg = 0.415 g

0.415 g / 194.19 g/mol = 2.14×10⁻³ mol

[Caffeine] = 2.14×10⁻³ M

Let's calculate pH. As we don't have Kb, we can obtain it from pKb.

- log Kb = pKb → 10^-pKb = Kb

10⁻¹⁰'⁴ = 3.98×10⁻¹¹

We go to equilibrium:

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

Initially we have 2.14×10⁻³ moles of caffeine, so, after the equilibrium we may have (2.14×10⁻³ - x)

X will be the amount of protonated caffeine and OH⁻

Caffeine + H₂O ⇄ Caffeine⁺ + OH⁻ Kb

(2.14×10⁻³ - x) x x

We make the expression for Kb:

3.98×10⁻¹¹ = x² / (2.14×10⁻³ - x)

We can missed the -x in denominator, because Kb it's a very small value.

So: 3.98×10⁻¹¹ = x² / 2.14×10⁻³

√(3.98×10⁻¹¹ . 2.14×10⁻³) = x → 2.92×10⁻⁷

That's the [OH⁻]. - log [OH⁻] = pOH

- log 2.92×10⁻⁷ = 6.53 → pOH

14 - pOH = pH → 14 - 6.53 = 7.47

4 0

3 years ago

Another question please help!!! :)

Oxana [17]

The answer is [Boron]

I had this answer on my Chemistry test!

6 0

3 years ago

Mr. Cortez asked his students to give examples in which electrical energy is transformed into heat. Students worked in a small g

Lunna [17]

You have to post the table

7 0

3 years ago

Read 2 more answers