How do you separate iron and steel from the industrial waste

A buret is filled with 0.1517 M A 25.0 mL portion of an unknown acid and two drops of indicator are added to an Erlenmeyer flask

erik [133]

Answer:

Molarity of Unknown Acid = 0.1332 M

Explanation:

Data for solving problem:

Molarity of base in buret (M₁)= 0.1517 M

volume of the acid in Erlenmeyer flask (V₂)= 25.0 mL

Volume of the base in the buret (V₁) = final volume of buret - initial volume in buret

final volume of buret = 22.5 mL

initial volume in buret = 0.55 mL

So

Volume of the base in the buret (V₁) = 22.5 mL -0.55 mL = 21.95 mL

Volume of the base in the buret (V₁) = 21.95 mL

Molarity of Unknown acid in the Erlenmeyer flask (M₂) = To be find

Explanation:

It is acid base titration and formula for this titration is as follows:

Molarity of base x Volume of base = Molarity of acid x volume of acid

it can be written as

M₁V₁ = M₂V₂ -------------------- equation (1)

we have to find M₂

so by rearrangment the equation (1)

M₁V₁ / V₂ = M₂ ------------------ equation (2)

put the values in equation in equation (2)

M₂ = 0.1517 M x 21.95 mL / 25.0 mL

M₂ = 3.3298 /25.0

M₂ = 0.1332 M

so the Molarity of Unknown acid is <u>0.1332 M</u>

5 0

3 years ago

How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?

SpyIntel [72]

Answer:

The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

$ln (\frac{C}{C_0})=-kt$

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as $5.11 \times 10^{-5} \, s^{-1}$

So time t is given as

$ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs$

So the concentration will become 15% of the initial value after 10.31 hrs.

7 0

2 years ago

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Which pair shares the same empirical formula?

ZanzabumX [31]

Answer:- 3. $CH_3$ and $C_2H_6$

Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.

For example, the molecular formula of benzene is $C_6H_6$ . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is $CH_2$ . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is $CH_2$ . The C to H ratio for second molecule is 1:4, so the empirical formula is $CH_4$ . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is $CH_3$ . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also $CH_3$ . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is $CH_2$ . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. $CH_3$ and $C_2H_6$

4 0

3 years ago

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Calculate the percent ionization of nitrous acid in a solution that is 0.139 M in nitrous acid. The acid dissociation constant o

anzhelika [568]

Ok first, we have to create a balanced equation for the dissolution of nitrous acid.

HNO2 <-> H(+) + NO2(-)

Next, create an ICE table

HNO2 <--> H+ NO2-
[]i 0.139M 0M 0M
Δ[] -x +x +x
[]f 0.139-x x x

Then, using the concentration equation, you get

4.5x10^-4 = [H+][NO2-]/[HNO2]

4.5x10^-4 = x*x / .139 - x

However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,

assume 0.139-x ≈ 0.139

4.5x10^-4 = x^2/0.139

Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.

We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.

Then to find percent dissociation, you do final concentration/initial concentration.

0.007M/0.139M = .0503 or

≈5.03% dissociation.

4 0

3 years ago

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Hydroelectric power is produced using the energy of moving water.

SIZIF [17.4K]

Answer:

it's da 3rd answer

Explanation:

3 0

3 years ago