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2 years ago

Over-the-counter hydrogen peroxide solutions are 3% (w/v). What is this concentration in moles per liter?

1 answer:

5 0

1 ml of water weighs 1 gram so 1 liter (1000 ml) weighs 1000 grams. A 3% solution (3% = 0.03) of hydrogen peroxide (w/v) would contain 1000 grams x 0.03 or 30 grams. The chemical formula of hydrogen peroxide is H2O2 and a mole weighs 34.0147 grams/mole. So 30 grams of H2O2 divided by 34.0147 grams/mole equals 0.88 moles of H2O2. The concentration of a 3% (w/v) hydrogen peroxide solution therefore contains 30 grams of H202 (or 0.88 moles of H202) per in a liter of water (or 1000 grams H20) would thus be 0.88 moles H2O2 per liter (0.88 moles H2O2/l) .

You might be interested in

What alkene reacts the fastest with HBr?

Rasek [7]

The first step in the reaction is the double bond of the Alkene going after the H of HBr. This protonates the Alkene via Markovnikov's rule, and forms a carbocation. The stability of this carbocation dictates the rate of the reaction.

So to solve your problem, protonate all your Alkenes following Markovnikov's rule, and then compare the relative stability of your resulting carbocations. Tertiary is more stable than secondary, so an Alkene that produces a tertiary carbocation reacts faster than an Alkene that produces a secondary carbocation.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

3 0

2 years ago

Read 2 more answers

If you find that the hydrogen alpha line in a star’s spectrum occurs at a wavelength of 656.45 nm, what is the star’s approximat

Paladinen [302]

Answer:

The answer to the question is

The star’s approximate radial velocity is 68.52 km/s

Explanation:

To solve the

The formula is

$\frac{\lambda-\lambda_{rw}}{\lambda_{rw}} = \frac{v_r}{c}$ where

$v_r$ = velocity of the star

λ = Star's spectrum wavelength = 656.45 nm

$\lambda_{rw}$ = Rest wavelength = 656.30 nm

c = Speed of light = 299 792 458 m / s

Therefore we have

$\frac{656.45-656.30}{656.30} =\frac{v_r}{299 792 458}$ or $v_r$ = 68518.7699 m/s or 68.52 km/s

8 0

2 years ago

A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i

almond37 [142]

The half reaction that occurs at the Au electrode is 1.64

Explanation:

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e- ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e- ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1 0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is 1.64

6 0

2 years ago

According to the following balanced reaction, how many moles of HNO3 are formed from 8.44 moles of NO2 if there is plenty of wat

kotegsom [21]

Answer:

5.63 mol.

Explanation:

The balanced chemical equation between NO₂ and H₂O is:

3NO₂(s) + H₂O(l) → 2HNO₃(aq) + NO(g),

It is clear that 3 mol of NO₂ reacts with 1 mol of H₂O to produce 2 mol of HNO₃ and 1 mol of NO.

Water is present as an excess reactant and NO₂ is limiting reactant.

To find the no. of moles of HNO₃ produced:

3 mol of NO₂ produces → 2 mol of HNO₃, from stichiometry.

8.44 mol of NO₂ produces → ??? mol of HNO₃.

∴ The no. of moles of HNO₃ are formed = (8.44 mol)(2 mol)/(3 mol) = 5.63 mol.

3 0

2 years ago

Which of the following concerning electrolytes and the solubility rules is/are true? 1. The solubility rules apply only to ionic

Mkey [24]

Answer:

1. is true

Explanation:

The solubility rules apply only to salts, which are ionic compounds.

2. is false. A strong electrolyte is a salt that dissociates completely in solution. Not all salts dissociate completely. For example, a 0.36 mol·L⁻¹ solution dissociates as:

K₂SO₄ ⟶ K⁺ + KSO₄⁻ (30 %) + SO₄²⁻

Thus, K₂SO₄ does not dissociate completely into K⁺ and SO₄²⁻ ions.

3. is false. The solubility rules apply only to aqueous solutions.

7 0

2 years ago