Answer:
I just guessed it out to be 3
Answer:
23.27 L
Explanation:
The balanced reaction of ammonia and nitrogen monoxide is shown below as:
4NH₃(g) + 6NO(g) ⇒ 5N₂(g) + 6H₂O(l)
Given that:
Volume of nitrogen monoxide = 34.9 L
Since, temperature and pressure are same, volume coefficients would be same as equation coefficients.
So,
6 L of nitrogen monoxide reacts with 4 L of ammonia.
1 L of nitrogen monoxide reacts with 4/6 L of ammonia.
34.9 L of nitrogen monoxide reacts with (4/6)*34.9 L of ammonia.
Amount of ammonia required = 23.27 L
The answer would have to be letter b. Energy
To determine the mass of xenon tetrafluoride, we need to know first the number of fluorine atoms present in <span>oxygen difluoride. We need to convert first the mass into moles then make use of the relation of the elements from the chemical formula. Then, use the avogadro's number to convert it to number of atoms. Then, we do the reverse of the steps above but this time for </span><span>xenon tetrafluoride.
25.0 g OF2 ( 1 mol / 54 g ) ( 2 mol F / 1 mol OF2 ) ( 6.022 x10^23 atoms F / 1 mol F ) ( 1 mol / 6.022x10^23 atoms) ( 1 mol XeF4 / 4 mol F ) (207.3 g / 1 mol XeF4) = 47.99 g XeF4</span>
