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iren2701 [21]
3 years ago
8

A solution with a ph of 4 has _________ the concentration of h+ present compared to a solution with a ph of 5.

Chemistry
1 answer:
Marina CMI [18]3 years ago
7 0
<span>A solution with a pH of 4 has ten times the concentration of H</span>⁺<span> present compared to a solution with a pH of 5.
</span>pH <span>is a numeric scale for the acidity or basicity of an aqueous solution. It is  the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
</span>[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.
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Bromine can be classified as a
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Answer:

Halogen / salt-former

Explanation:

Bromine is classified as an element in the 'Halogens' section which can be located in group 7 of the Periodic Table. The term "halogen" means "salt-former" and compounds containing halogens are called "salts".

8 0
3 years ago
One glass of water is 85 degrees F and another is 40 degrees F. When an Alka Seltzer tablet is dropped into each glass, at the s
Andreyy89

Answer:

the cold will get hot and the hot will get cold

Explanation:

7 0
2 years ago
A gas occupies 3.5L at 2.5 kPa pressure. What is the volume at 100 mmHg at the same temperature? Be sure to
nataly862011 [7]

Answer:

V₂ = 0.656 L

Explanation:

Given data:

Initial volume = 3.5 L

Initial pressure = 2.5 KPa

Final volume = ?

Final pressure = 100 mmHg (100/7.501=13.33 KPa)

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

2.5 KPa × 3.5 L = 13.33 KPa × V₂

V₂ = 8.75 KPa. L/13.33 KPa

V₂ = 0.656 L

7 0
2 years ago
Annabelle was explaining the carbon cycle to her friend. She said that all the carbon
vagabundo [1.1K]

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I DON'T THINK SHE IS CORRECT

Explanation:

5 0
3 years ago
If you have 12.5g of fluoride and 16.2g of sodium, which is the limiting reactant and how sodium fluoride in grams is your theor
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Answer:

F2 is the limiting reactant

27.6 grams of NaF is produced.

Explanation:

Balance the equation first.

2Na+ F2 ---> 2NaF

To find the limiting reactant, solve for how much NaF can be produced with Na and F2

12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)

=0.658 moles NaF

16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)

=0.705 moles NaF

Since F2 produced the least NaF, F2 is the limiting reactant.

Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.

0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF

27.6 moles of NaF would be theoretically produced.

8 0
3 years ago
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