Can u pls help me naming this: (IUPAC)

In a titration experiment, 31.4 mL of 1.120 M HCOOH is neutralized by 16.3 mL of Ba(OH)2. What is the concentration of the Ba(OH

11Alexandr11 [23.1K]

Answer : The concentration of the $Ba(OH)_2$ solution is, 2.16 M

Explanation :

To calculate the concentration of base, 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 $HCOOH$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$ .

We are given:

$n_1=1\\M_1=1.120M\\V_1=31.4mL\\n_2=2\\M_2=?\\V_2=16.3mL$

Putting values in above equation, we get:

$1\times 1.120M\times 31.4mL=2\times M_2\times 16.3mL\\\\M_2=2.16M$

Thus, the concentration of the $Ba(OH)_2$ solution is, 2.16 M

7 0

3 years ago

In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.

mamaluj [8]

Answer:

D

Explanation:

We know that the

reaction catalyzing power of a catalyst ∝ surface area exposed by it

Given

volume V1= 10 cm^3

⇒ $\frac{4}{3} \pi r^3= 10$

hence r= 1.545 cm

also, surface area S1= $4\pi r^2$

now when the sphere is broken down into 8 smaller spheres

S2= 8×4πr'^2

now, equating V1 and V2 ( as the volume must remain same )

$\frac{4}{3}\pi r^3=8\times\frac{4}{3} \pi r'^3$

and solving we get

r'= r/2

therefore, S2= $8\times4\pi\frac{r}{2}^2$

S2= $2\times4\pi r^2$

S2= 2S1

hence the correct answer is

. The second run has twice the surface area.

8 0

3 years ago

which Substance can be used to Neutralize nitric acid? A Acetic acid C Hydrogen peroxide b milk of magnesia D Ethanol

sweet-ann [11.9K]

Answer:B

Explanation:

7 0

2 years ago

The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2

Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K = 6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0

3 years ago

Elements that belong to the same period/row!!

-BARSIC- [3]

Answer:

Each row is called a period. Each column is called a group or family.

Explanation:

7 0

3 years ago