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stiv31 [10]
4 years ago
13

If I have 24.25 grams of nitrogen, how many moles is that?

Chemistry
1 answer:
Hunter-Best [27]4 years ago
6 0
I got a wayyyy longer decimal so I just shortened it to 1.73.
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What are hydroxyl groups and what happens to them when molecules combine?
Sloan [31]
Hi! Chemistry is complicated, right?

A hydroxyl group is a group that consists of
oxygen and hydrogen bonds. When they combine, ethanol may be released.

I hope I helped!
5 0
3 years ago
Which of the following is the most electronegative element?
Kipish [7]
None of these; the most electronegative element is fluorine because an atom of fluorine needs an additional electron to join it's outer membrane, so the atom can achieve stability.

a flourine atom on its own is always a negative ion.


7 0
3 years ago
Using the potential given below, calculate the activity of cl^- in 1M KCl solution.
MissTica

Answer:

Explanation:

E° (calomel electrode) = 0.268 v

E( calomel electrode, 1M KCl) = 0.280 v

For calomel electrode

E = E^0 _( Hg^+/Hg ) - \frac{.059}{2} \times log( a_{cl^-} )^2

.280 = .268 - \frac{.059}{2} \times log( a_{cl^-} )^2

.012=  - \frac{.059}{2} \times log( a_{cl^-} )^2

log( a_{cl^-})^2= - .4067

a_{cl^-_}  = 0.626 .

3 0
3 years ago
Nitrogen gas is being held in a 14.3 L tank at a temperature of 62°C. What will the volume be when the temperature drops to 24°C
Mice21 [21]
I think the volume should be still 14.3L. In this question, the gas is being held in a tank. The tank has a constant volume. The gas in the tank has the property to fill the space. So when the temperature changes, the pressure changes too. But the volume doesn't change.
5 0
3 years ago
Calculate the concentration (M) of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to
statuscvo [17]

Answer:

Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M

≈0.350 M

Explanation:

From the question it is clear that,  

Initial volume of sodium sulphide solution is (v₁) = 50mL

Initial concentration of sodium sulphide solution is (s₁) =0.874 M

Final volume of sodium sulphide solution is (v₂) = 250mL

Let, the final concentration of sodium sulphide solution is s₂, then according to acidimetry-alkalimetry,

v₁ × s₁ = v₂ × s₂

Or, s₂ = v₁ × s₁/v₂

= 50 × 0.874 / 250

= 0.1748 M

Therefore, concentration of 250mL sodium sulphide solution is 0.1748 M

Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M

≈0.350 M

5 0
3 years ago
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