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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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How many grams of sodium chloride are in 250mL of a 2.5M NaCI solution
Elden [556K]
I think I did this right.

4 0
4 years ago
Which sound wave measures loudness?
Kaylis [27]
The answer would be C. Amplitude
7 0
4 years ago
2. What are the uncommon states of matter?
mixer [17]
Most uncommon states of matter:
c. Superfluid, plasma


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8 0
3 years ago
In the reaction ca + f2 caf2 , what should be the coefficient for f2?
alex41 [277]
Answer : In the reaction of Ca + F _{2} --->CaF _{2}, the coefficient of Ca and F _{2} will be the same which will be 1.

Explanation : As per the rules of writing a chemical equation and balancing it properly, the number of elements in reactants side should be equal to the number of elements generated in product side. Hence when we use 1 molecule of Ca and one molecule of F _{2} then it generates 1 molecule of CaF _{2}.
6 0
4 years ago
If 40.0 g of molten iron(II) oxide reacts with 10.0 g of mag-nesium, what is the mass of iron produced
vagabundo [1.1K]

Answer:

m_{Fe}=23.0gFe

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

FeO+Mg\rightarrow Fe+MgO

Thus, for the given masses of reactants we should compute the limiting reactant for which we first compute the available moles of iron (II) oxide:

n_{FeO}=40.0gFeO*\frac{1molFeO}{72gFeO} =0.556molFeO

Next, we compute the consumed moles of iron (II) oxide by the 10.0 g of magnesium, considering their 1:1 molar ratio in the chemical reaction:

n_{FeO}^{consumed}=10.0Mg*\frac{1molMg}{24.3gMg}*\frac{1molFeO}{1molMg}=0.412molFeO

Therefore, we can notice there is less consumed iron (II) oxide than available for which it is in excess whereas magnesium is the limiting reactant. In such a way, the produced mass of iron turns out:

m_{Fe}=0.412molFeO*\frac{1molFe}{1molFeO}*\frac{56gFe}{1molFe}\\  \\m_{Fe}=23.0gFe

Regards.

8 0
4 years ago
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