All categories

3 years ago

The pH of a solution is measured as 8.3. What is the hydrogen ion concentration of the solution?

Chemistry

1 answer:

castortr0y [4]3 years ago

8 0

The logarithmic of the reciprocal of hydrogen - ion concentrate in grams atoms per litre

You might be interested in

Give an example of a non-biological chemical reaction that involves carbon. Name the process that converts a solid compound into

xxMikexx [17]

Answer:

The process that converts a solid mcompound into a gas is sublimation.

E.g- dry ice, solid iodine and ammonium salts

when the above solid are heated, only particles which are found on the surface of the solid gain enough energy and break all forces of attraction and form a gas.

Explanation:

6 0

3 years ago

How can atoms be neutral if they contain charged particles?

rewona [7]

Stability of atoms is determined by neutron:proton ratio. This n/p ratio is 1:1 for elements below atomic number 20. Hope this helps.

4 0

3 years ago

A 2.36-gram sample of NaHCO3 was completely decomposed in an experiment.

enyata [817]

Answer:

Explanation:

a) The mass of the reactants is 2.36 grams, and the mass of the products is 1.57 grams plus the mass of the carbonic acid. Thus, using the law of conservation of mass, we get the mass of the carbonic acid is 2.36 - 1.57 = 0.79 grams.

b) The gram-formula mass of sodium bicarbonate is 84.006 g/mol, meaning that 2.36/84.006 = 0.028 moles were consumed. Thus, this means that in theory, 0.014 moles of carbonic acid should have been produced, which would have a mass of (0.014)(62.024)=0.868 grams. Thus, the percentage yield is (0.79)/(0.868) * 100 = 91%

8 0

2 years ago

A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°C. What is the magnitude of k at 95.0°C if Ea =

never [62]

Answer:

k ≈ 9,56x10³ s⁻¹

Explanation:

It is possible to solve this question using Arrhenius formula:

$ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )$

Where:

k1: 1,35x10² s⁻¹

T1: 25,0°C + 273,15 = 298,15K

Ea = 55,5 kJ/mol

R = 8,314472x10⁻³ kJ/molK

k2 : ???

T2: 95,0°C+ 273,15K = 368,15K

Solving:

$ln\frac{k2}{k1} = 4,257$

$\frac{k2}{k1} = 70,593$

${k2} = 9,53x10^3 s^{-1}$

k ≈ 9,56x10³ s⁻¹

I hope it helps!

5 0

3 years ago

Suppose that a catalyst lowers the activation barrier of a reaction from 125kJ/mol to 55kJ/mol125⁢kJ/mol to 55kJ/mol. By what fa

sineoko [7]

Answer:

The factor of increasing reaction rate is 1,85x10¹².

Explanation:

Using arrhenius formula:

$k = A e^\frac{-E_{a}}{RT}$

Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K

Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:

k = 1,25x10⁻²²

When activation energy is 55kJ/mol:

k = 2,31x10⁻¹⁰

Thus, the factor of increasing reaction rate is:

2,31x10⁻¹⁰/1,25x10⁻²² = 1,85x10¹²

I hope it helps!

8 0

3 years ago