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irina [24]
3 years ago
14

What products will form when potassium is added to cold, liquid water?

Chemistry
2 answers:
kow [346]3 years ago
7 0

Answer:

The answer is C

Explanation:

i took a test on this.

rewona [7]3 years ago
4 0

Answer:

C

Explanation:

Potassium metal reacts very rapidly with water, which forms Potassium hydroxide (KOH) and hydrogen gas (H2).

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A. 1.4 m<br> B.1.9m<br> C. 2.4m<br> D. 3.6m
jenyasd209 [6]
Apple the way to my hair and hair is so good and i and my friend are going to sleep at the house and i you can do to me and you can do it and i and you have a lot to do with me some time i to get a little sleep in my room and you have my sleep ritirieuwi u and the only reason
8 0
2 years ago
Write the net ionic equation for the reaction that occurs when a solution of hydrochloric acid (hcl) is mixed with a solution of
kap26 [50]
Ionic Equation:
H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq) + Na⁺(aq) + Cl⁻(aq)

Net ionic equation:
H⁺(aq) + CHO₂⁻(aq) → HCHO₂(aq)
5 0
3 years ago
Read 2 more answers
Please help me:) thanks!
Cloud [144]

Answer:

2Cl2 has 4 atoms

NaOH has 3 atoms

CaCO3 has 5 atoms

Explanation:

4 0
2 years ago
Read 2 more answers
Chlorine is a very reactive non-metals. Why
Vadim26 [7]

Answer:

Chlorine is a very reactive non metals because this element don't form any known chemical compound.

Explanation:

Chlorine atoms have a lot of valance electrons without being complete on its own, so it has a greater need to seek it's conjugates. That's called electronegativity. Halogens are highly reactive because of their electronegativity.

3 0
2 years ago
If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? T
Veronika [31]

Answer:

m_{MnO_2}=21.2gMnO_2

Explanation:

Hello,

In this case, given the balanced reaction:

2MnO_4^-+2Mn(OH)_2\rightarrow 4MnO_2+2OH^-+H_2O

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

m_{MnO_2}=14.5gMnO_4^-*\frac{1molMnO_4^-}{118.9gMnO_4^-}*\frac{4MnO_2}{2molMnO_4^-}  *\frac{86.9gMnO_2}{1molMnO_2} \\\\m_{MnO_2}=21.2gMnO_2

Best regards.

5 0
3 years ago
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