All categories

2 years ago

What volume (mL) of a 0.3428 M HCl(aq) solution is required to completely neutralize 23.55 mL of a 0.2350 M Ba(OH)2(aq) solution

Chemistry

1 answer:

EastWind [94]2 years ago

6 0

Answer:

The volume required for complete neutralize is 32.29 mL

Explanation:

The computation of the volume required for complete neutralize is shown below:

As we know that, the balanced equation is

$Ba(OH)_2 + 2Hcl\rightarrow Bacl_2 + 2H_2O$

Now

The number of moles of $Ba(OH)_2$ = n_1 = 1

And, the number of moles of Hcl = n_2 = 2

Therefore

The equation i.e. to be used to find out the volume is given below:

$\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$

$V_2 = \frac{M_1V_1}{n_1} \times \frac{n_2}{M_2} \\\\ = \frac{0.2350 \times 23.55}{1} \times \frac{2}{0.3428} \\\\ = \frac{11.0685}{0.3428}$

= 32.29 mL

Hence, the volume is 32.29mL

You might be interested in

According to reference table adv-10, which reaction will take place spontaneously?

olga_2 [115]

Missing table!! write the elements with the first letter of the symbol with Upper Caps letters!!!

http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm

Ni2+ +Pb(s) → Ni(s) + Pb2+
The potential of the oxidation of Pb(s) --> Pb2+(aq) is 0.126 V
The potential of the reduction go Ni2+(aq) --> Ni(s) is -0.25 V

Add the two together and the potential for the reaction is -0.124 V (NO SPONTANEOUS THE SIGN IS NEGATIVE)

au3+ + al(s) → au(s) + al3+Au3+(aq) -> Au(s) +1.5 VAl -> Al3+ +1.66VV= 3.16 (SPONTANEOUS THE SIGN OF THE PONTENTIAL IS POSITIVE)Sr2+ + Sn(s) → Sr(s) + Sn2+

Sr2+(aq) + 2 e– Sr(s) V= -2.89V
Sn -> Sn2+ V= 0.14 V
V= -2.75 V (no spontaneous)

Fe2+ + Cu(s) → Fe(s) + Cu2+
Fe2+(aq) + 2 e– Fe(s) V= -0.44 V
Cu -> C2+ V = - 0.337V

V= - 0.777V (no spontaneous)

5 0

2 years ago

A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete

lys-0071 [83]

ke prop to v^2

ke1/v1^2=ke2/v2^2

400/50x50=joules/100x100

400x2x2

1600j

8 0

2 years ago

A deck of cards has dimensions 8.9cm x 5.72cm x 1.82 cm. What is the volume of the deck in cubic centimeters?

arsen [322]

Answer:

92.65256 cm^3

Explanation:

To find this, we can simply multiply all three dimensions to get the answer in cubic centimeters, and we get the answer above. If you want to be more specific, we can go by the sigfig rule and the answer would be rounded to 93 cm^3.

5 0

2 years ago

Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0

ExtremeBDS [4]

Answer:

The freezing point will be $2.9^{0}C$

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

$Depressioninfreezing point=K_{f}Xmolality$

Where

Kf= $20\frac{^{0}C}{m}$

the molality is calculated as:

$molality=\frac{moles_{solute}}{mass_{solvent}}$

$moles=\frac{mass}{molarmass}=\frac{0.694}{154}=0.0045mol$

$massofcyclohexane=25g=0.025Kg$

$molarity=\frac{0.0045}{0.025}=0.18m$

Depression in freezing point = $20X0.18=3.6^{0}C$

The new freezing point = $6.5^{0}C-3.6^{0}C=2.9^{0}C$

5 0

2 years ago

The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T

kifflom [539]

Answer : The equilibrium $SO_2$ pressure is, $3.93\times 10^{-5}torr$

Explanation :

The given balanced chemical reaction is,

$SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)$

First we have to calculate the standard free energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]$