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:
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:
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
Answer: 0.405g
Explanation:
Molar Mass of Be = 9g/mol
Number of mole of Be = 0.045mol
Mass conc. Of Be = 0.045 x 9 = 0.405g
Answer:
6.43 moles of NF₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.
Thus, 6.43 moles of NF₃ were obtained from the reaction.
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