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Romashka [77]
3 years ago
8

How many moles of electrons are required to reduce one mole of nitrogen gas (N2) to two moles of nitrogen ions (N3-)?

Chemistry
1 answer:
Makovka662 [10]3 years ago
8 0
Start by writing the atoms balance:

N_{2} -\ \textgreater \  2N^{3-}

Now, determine the change of oxidation states:N_{2} has oxidation state 0, so each N has to gain 3 electrons to become N^{3-}.

That, means that you need 6 electrons to balance the charges, resulting in:

N_{2} + 6  e^{-} -\ \textgreater \  2 N_{3-}

And the answer is 6 mole of electrons.
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Minerals are _________ inorganic _________ that usually possess a crystalline structure and can be represented by a chemical for
ladessa [460]

Answer:

The corect answer is c) naturally occurring; solids

Explanation:

Minerals exists as solid substances in nature consisting of one or more element chemically combined together formiming compounds with definite composition. As mentioned earlier single elements can form minerals and  examples of single element mineral are Silver, Carbon and Gold which are found in nature in their pure form and are mined.

Minerals are normally found in rocks, which may contain one ore more different types of minerals

7 0
3 years ago
Zn + O2= ZnO<br><br> How many moles of zinc are needed to make 500. g of zinc oxide?​
s2008m [1.1K]

Answer:

5.15 moles

Explanation:

2zn + o2 = 2zno

5.15 2.57 5.15 moles

nzno=500/(16x2+65)= 5.15 moles

-> nzn = 5.15 x 2 ÷ 2 = 5.15 moles

6 0
3 years ago
Is this right, I feel like its wrong
hjlf

It is correct, next time re-check your answer and don't second guess yourself. ;3

7 0
3 years ago
What is the correct formula
nataly862011 [7]

Answer:

A the answer is A I'm sure

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3 years ago
The ΔHcomb value for anethole is -5539 kJ/mol. Assume 0.840 g of anethole is combusted in a calorimeter whose heat capacity (Cal
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Answer:

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Explanation:

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So, 0.840 g of anethole = \frac{0.840}{148.2}moles of anethole = 0.00567 moles of anethole

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So, 0.00567 moles of anethole release (5539\times 0.00567)kJ of heat or 31.41 kJ of heat

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So, 31.41 kJ of heat increases (\frac{1}{6.60}\times 31.41)^{0}\textrm{C} or 4.76^{0}\textrm{C} temperature of calorimeter

So, the final temperature of calorimeter = (20.6+4.76)^{0}\textrm{C}=25.36^{0}\textrm{C}

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3 years ago
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