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zhannawk [14.2K]
3 years ago
6

H2(g) + F2(g) → 2 HF(g) ΔH=-546 kJ

Chemistry
2 answers:
Ulleksa [173]3 years ago
5 0

Answer: The answer is D, More energy was released in the formation of the 2 HF bonds than in the breaking of the bonds of H2 and F2.

Explanation: Honestly, I guessed on this question and got it right, but I see that right after the arrow on the right side of the equation, there's a 2 in front of the HF(g). On the left side of the equation, it does not match the right side, which led me to believe that there was more energy being released.

Arisa [49]3 years ago
4 0

Answer: help please !!

Explanation:

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please I think the answer is A.

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5 0
3 years ago
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The ratio of nitrogen to oxygen by mass in NO is 7.0:8.0. Identify the ratio of nitrogen to oxygen by mass in NO2 and N2O7 .
Rus_ich [418]

This problem is providing the ratio of nitrogen to oxygen by mass in nitrogen monoxide, NO, as 7.0:8.0 and asks for the same ratio but in NO₂ and N₂O₇. After doing the calculations, the results are 7.0:16.0 and 1.0:4.0 respectively.

<h3>Mass ratios:</h3>

In chemistry, one can calculate the mass ratios in chemical formulas according to the atomic mass of each atom. In such a way, one knows the mass ratio of nitrogen to oxygen in NO is 7.0:8.0 because we divide the atomic mass of nitrogen by oxygens:

\frac{14}{16}=\frac{7.0}{8.0}

Now, for chemical formulas with subscripts, one must multiply the atomic mass of the element by the subscript in the formula, which is the case of NO₂ and N₂O₇ as shown below:

NO_2:\frac{14}{16*2}=\frac{14}{32} =\frac{7.0}{16.0}  \\&#10;\\&#10;N_2O_7:\frac{14*2}{16*7}=\frac{28}{112} =\frac{1.0}{4.0}

Therefore, the results for NO₂ and N₂O₇ are 7.0:16.0 and 1.0:4.0 respectively

Learn more about atomic masses: brainly.com/question/5566317

3 0
2 years ago
Abbreviation for mole
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Answer:

Maybe mol

Explanation:

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3 years ago
10 POINTS
Snowcat [4.5K]
E=hc/λ =6.626×10^-34×3 ×10^8 / 3×10^7 × 10^-9 = 6.626×10 ^-24J.
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3 years ago
[Rn] 7s25f6d4<br> Which element is denoted
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Answer:

Explanation:

[Rn] 7s25f6d4

*-*

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