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sasho [114]
3 years ago
5

A sample of juice has a pH of 4.2. Whsat is the concentration in mol/dm3 of OH in the juice?​

Chemistry
1 answer:
Annette [7]3 years ago
6 0

Answer:

6 * 10^-4M

Explanation:

use this equation

[ H + ]  =  10  −  pH

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In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for elect
qwelly [4]

Answer:

The 1st and 4th options are correct

I.the oxidized form has a higher affinity for electrons

IV. the greater the tendency for the oxidized form to accept electrons

Explanation:

Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.

Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.

(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.

4 0
3 years ago
Which highly reactive gas was probably absent from the Earth's primitive atmosphere? O2 (oxygen gas) water vapor methane carbon
rosijanka [135]
Oxygen gas was most likely absent from Earth's primitive atmosphere.  The current theory is that the Earth's early atmosphere was composed of mainly carbon dioxide and methane due to the high volcanic activity.  Cyanobacteria and their use of photosynthesis was what caused earth's atmosphere to become oxygen enriched.
I hope that helps.
6 0
3 years ago
I need to repost cause I did it wrong and the price was blank, but imma mark BRAINLIEST on this one :D
k0ka [10]

Explanation:

I'm pretty sure 1. yes 2. no and 3. it might be yes but I'd just put a maybe

5 0
3 years ago
You will need to prepare 12 mL of 25% Sodium Phosphate Buffer (pH 4) solution for Activity 2. What volume of the stock Sodium Ph
zhuklara [117]

Assuming the concentration of stock solution is 50% sodium phosphate buffer solution, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

<h3>What volume of a stock Sodium phosphate buffer and water is needed to 12 mL of 25% sodium phosphate buffer of pH 4?</h3>

The process of preparing solutions from stock solutions of higher concentration is known as dilution.

Dilution is done with the aid of the dilution formula given below:

  • C1V1 = C2V2

where

  • C1 is the concentration of stock solution
  • V1 is the volume of stock solution required to prepare a diluted solution
  • C2 is the concentration of the diluted solution prepared
  • V2 is the final volume of the diluted solution

From the data provided:

C1 is not given

V1 is unknown

C2 = 25%

V2 = 12 mL

  • Assuming C1 is 50% solution

Volume of stock, V1, required is calculated as follows:

V1 = C2V2/C1

V1 = 25 × 12 /50

V1 = 6 mL

Therefore, the volume of stock solution required is 6 mL and the volume of water required is 6 mL.

Learn more about dilution formula at: brainly.com/question/7208546

6 0
2 years ago
Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen
andrew-mc [135]
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O 
the limiting reactant in this reaction is C₄H₁₀  This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
7 0
3 years ago
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