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GenaCL600 [577]
3 years ago
9

Groups of elements that are shiny but not lustrous, have a semi-conductive property, and are brittle are classified as?

Chemistry
1 answer:
9966 [12]3 years ago
3 0
Answer:
A. Nonmetals
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Please someone solve this and tell me how you solve it
lianna [129]

Answer:

Supersaturated.

Explanation:

Hello there!

In this case, according to this solubility chart, we infer that for NH3, the solubility starts at 90 grams of NH3 that are soluble in 100 g of water at 0 °C and ends in about 8 g in 100 g of water at 100 °C for a saturated solution.

However, since we are asked for the solubility of NH3 at 20 °C, we can see that, according to the table and the curve for NH3, about 52 g of NH3 are soluble in 100 g of water; thus, for the given 60 g of NH3, we will say that 8 grams will remain undissolved, and therefore, this solution will be supersaturated.

Regards!

8 0
3 years ago
Earth's Revolution question:
Murljashka [212]

Answer:

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7 0
3 years ago
IF UR GOOD AT CHEM PLZ ANSWER MY PREVIOUS QUESTIONS ON MY PROFILE, I NEED HELP!!
erastova [34]

Answer:

Ok, I'll try

Explanation:

7 0
2 years ago
A compound's properties are different than the properties of the elements that make it up.
yan [13]
The answer is True because elements in a compound combine and become an entirely different substance with its own unique properties.
7 0
3 years ago
Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t
Veseljchak [2.6K]

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

4NH_3+7O_2\rightarrow 4NO_2+6H_2O

The expression for rate of reaction for the reactant :

\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}

\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}

The expression for rate of reaction for the product :

\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}

From this we conclude that, all the options are correct.

3 0
3 years ago
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