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Svetllana [295]
3 years ago
11

Help, cant get it. Answers in picture

Chemistry
1 answer:
Aleks04 [339]3 years ago
8 0

Answer:17.0

Explanation:

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What is true of a covalent bond? (03.03)
garri49 [273]

Answer:

I think B..

Explanation:

It is the sharing of electrons from one atom to another .

6 0
3 years ago
Read 2 more answers
Match the term with its description. (4 points)
Ne4ueva [31]

Answer:

A. a

B. c

C. b

D. d

Explanation:

3 0
3 years ago
What is the theoretical yield of Li3N in grams when 12.8 g of Li is heated with 34.9 g of N2?
Olenka [21]

Answer:- 21.4 grams of Li_3N are formed.

Solution:- The balanced equation is:

6Li+N_2\rightarrow 2Li_3N

From this equation, lithium and nitrogen reacts in 6:1 mol ratio. Limiting reactant gives the theoretical yield of the product. We will calculate the grams of the product for the given grams of both the reactants and see which one of them gives the limited amount of the product. This limited amount of the product will be the theoretical yield.

The molar mass of Li is 6.94 gram per mol and for N_2 It is 28.02 gram per mol. The molar mass of Li_3N is 34.83 gram per mol. The calculations for the grams of the product for given grams of both the reactants are shown below:

12.8gLi(\frac{1molLi}{6.94gLi})(\frac{2molLi_3N}{6molLi})(\frac{34.83gLi_3N}{1molLi_3N})

= 21.4gLi_3N

34.9gN_2(\frac{1molN_2}{28.02gN_2})(\frac{2molLi_3N}{1molN_2})(\frac{34.83gLi_3N}{1molLi_3N})

= 86.8gLi_3N

From above calculations, Li gives least amount of the product. So, 21.4 g of Li_3N are formed.

4 0
3 years ago
What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?
zubka84 [21]
Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g 
H: 16.3% = 16.3 g 

Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

C:  \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol

H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

C:  \dfrac{6.975}{6.975} = 1

H:  \dfrac{16.3}{6.975} \approx 2.3

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark

I hope this helps. =)
8 0
3 years ago
How do the atmosphere conditions near the beginning of Precambrian time contrast with the atmosphere conditions that are present
babunello [35]
The early precambrian atmosphere consisted primarily of nitrogen and carbon dioxide with almost no oxygen. 

<span>Today, the atmosphere contains about 20% oxygen, less carbon dioxide and similar amounts of nitrogen. </span>

<span>Photosynthetic green-leaf plants and trees are largely responsible for the change, converting carbon dioxide to oxygen.</span>
5 0
3 years ago
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