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Yanka [14]
3 years ago
9

Is it possible for a solution to have negative pH values?

Chemistry
2 answers:
aleksandr82 [10.1K]3 years ago
4 0

Answer:

Yes.

Explanation:

iogann1982 [59]3 years ago
4 0
Yes it is possible
Explanation:
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What’s the choices? Cuz it doesn’t make sense
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Consider the reaction cacn2 3 h2o → caco3 2 nh3 . how much nh3 is produced if 187 g of caco3 are produced?
shusha [124]
B ase from the reaction <span>cacn2 3 h2o → caco3 2 nh3, for every 1 mole of caco3 produced there 2 moles of nh3 being produced. to solved this, we must first convert the caco3 to moles.

mass nh3 = 187 g caco3 (1 mol caco3 / 100 g caco3 ) ( 2 mol nh3 / 1 mol caco3) ( 17 g nh3 / 1 mol nh3)

mass nh3 = 63.58 g nh3 is produced</span>
7 0
3 years ago
How does shielding impact the strength of attraction between the nucleus and the valence electrons?
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Ans: The shielding effect explains why valence shell electrons are more easily removed from the atom. The nucleus can pull the valence shell in tighter when the attraction is strong and less tight when the attraction is weakened. The more shielding that occurs, the further the valence shell can spread out.
5 0
4 years ago
What kind of properties can only be observed when a substance changes into a different substance? physical properties reaction p
Vitek1552 [10]

When a substance changes into a different substance, you are observing its chemical properties.

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8 0
3 years ago
Read 2 more answers
During lab, a student used a Mohr pipet to add the following solutions into a 25 mL volumetric flask. They calculated the final
kompoz [17]

Answer:

(FeSCN⁺²) = 0.11 mM

Explanation:

Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2

M (Fe(NO₃)₃  = 0.200 M

V (Fe(NO₃)₃ =  10.63 mL

n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol

M (KSCN) =  0.00200 M

V (KSCN) = 1.42 mL

n (KSCN) =  0.00200 * 1.42 = 0.00284 mmol

Total volume = V (Fe(NO₃)₃  + V (KSCN)

                       = 10.63 + 1.42

                       = 12.05 mL

Limiting reactant = KSCN

So,

FeSCN⁺² = 0.00284 mmol

M (FeSCN⁺²) = 0.00284/12.05

                     = 0.000236 M

Excess reactant = (Fe(NO₃)₃

n(Fe(NO₃)₃ =  2.126 mmol -  0.00284 mmol

                  =2.123 mmol

For standard 2:

n (FeSCN⁺²) = 0.000236 * 4.63

                    =0.00109

V(standard 2) = 4.63 + 5.17

                       = 9.8 mL

M (FeSCN⁺²)  = 0.00109/9.8

                      = 0.000111 M = 0.11 mM

Therefore, (FeSCN⁺²) = 0.11 mM

7 0
3 years ago
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