A device that transforms chemical energy to electrical energy, and provides electrical force in a current.

pls help!!! Calculate the pH of a solution with a H+ ion concentration of 2.6 x 10^-7 and round your answer to the nearest hundr

earnstyle [38]

Answer:

6.59

Explanation:

pH=-log[H+]

=-log[2.6×10^-7]

=6.59

8 0

2 years ago

the half of an oxidation-reduction reaction shows the iron gains. what does this electron gain mean for iron.

Nataly [62]

The correct answer is option C, that is, it is reduced.

In reduction and oxidation reactions, reduction refers to the loss of an oxygen atom from a molecule or the gaining of one or more electrons. A reduction reaction is observed from the perspective of the molecule being reduced, as when one molecule gets reduced, another one gets oxidized. The complete reaction is called a redox reaction.

In the given case, iron gains electrons mean that it is reduced.

3 0

2 years ago

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentrati

Katen [24]

Answer:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

Explanation:

The ionization equation is

$H_{2}PO_{4}^{-} +H_{2}O$ ⇄ $HPO_{4}^{-2} +H_{3}O^{+}$ (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

$Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}$

The pKa is

$-Log (Ka) = -Log (6.2x10^{-8}) = 7.2$

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.

3 0

3 years ago

In the combustion of ethane, what is/are the reactant(s)?

aleksley [76]

Reactant are those that combine or reacts to give products !!

so in combustion of ethane ; ethane and o2 are reactants so ... your answer is B !!

5 0

3 years ago

Read 2 more answers

Suppose you have a 2.0 molar solution of sodium hydroxide (naoh), and you need 65 ml of 0.6 molar naoh. how should you make this

nalin [4]

The given molarity of sodium hydroxide solution = 2.0 M

The required concentration of sodium hydroxide is 65 mL of 0.6 M NaOH

Converting 65 mL to L:

$65mL*\frac{1L}{1000mL} =0.065L$

Calculating the moles of NaOH in the final solution:

$0.065L * \frac{0.6 mol}{L} =0.039mol$

Finding out the volume of 2.0 M solution taken to prepare the final solution:

$0.039 mol * \frac{1L}{2.0mol}=0.0195L*\frac{1000mL}{1L} =19.5mL$

Therefore, 19.5 mL of 2.0 M NaOH solution and make it up to 65 mL to prepare 0.6 M NaOH solution.

3 0

2 years ago