Answer: The empirical formula of one that contains 30.45% nitrogen is .
Explanation:
Given: Mass of nitrogen = 30.45 g
Let us assume that the mass of given oxide is 100 grams.
As the atomic mass of nitrogen is 14.0067 g. So, moles of nitrogen will be calculated as follows.
Also, mass of oxygen = (100 - 30.45) g = 69.55 g
Atomic mass of oxygen is 15.9994 g/mol. So, moles of oxygen will be as follows.
The ratio of both the atoms is as follows.
This means that gas has 2 moles of oxygen to 1 mole of nitrogen. Hence, the formula of oxide is .
Thus, we can conclude that the empirical formula of one that contains 30.45% nitrogen is .
Answer:
63.6%
Explanation:
The given compound is:
N₂O;
The problem here is to find the percent composition of nitrogen in the compound.
First find the molar mass of the compound:
Molar mass of N₂O = 2(14) + 16 = 44g/mol
So;
Percentage composition of Nitrogen = x 100 = 63.6%
Answer:
The final temperature of water is <u>20.5061 °C.</u>
Explanation:
Let the final temperature of water be 'x'.
Given:
Heat added to water is,
Initial temperature of water is,
Mass of water is,
Now, heat is added to water and its temperature is increased. The temperature is increased because water absorbs all the heat.
Heat absorbed by water is given as:
where 'c' is specific heat capacity of water and its value is equal to 4.186 J/g °C.
Now, plug in the given values and simplify.
Now, from law of conservation of energy, we know that:
Heat absorbed by water = Heat added to water
So, the final temperature of water is 20.5061 °C.
Answer:
Yeild of CO2 is approximately 54g
Explanation:
Using reaction stoichiomety and coeeficients, and knowing O2 is limiting reactant, 54 g of CO2 is produced.
Answer:
Methods for determining or delivering precise volumes include volumetric pipets and pycnometers; less precise methods include burets, graduated cylinders, and graduated pipets. In this experiment, you will measure masses and volumes to determine density. Four different metal cylinders are investigated.
Explanation: