Answer:
17.57kg of
and its percentage yield is 81.0%
Explanation:
Through the reaction you can get the theoretical amount of
that must be produced.

If the amount obtained is less than the theoretical amount, it means that the initial sample was not 100% pure. Now the actual amount obtained is compared with the theoretical amount using a percentage
=81.0%
<u>Answer:</u> The number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
<u>Explanation:</u>
In a chemical equation, the chemical species are termed as reactants or products.
Reactants are defined as the species which react in the reaction and are written on the left side of the reaction arrow.
Products are defined as the species which are produced in the reaction and are written on the right side of the reaction arrow.
For the given chemical equation:

On the reactant side:
Number of carbon atoms = (6 × 2) = 12
Number of hydrogen atoms = (14 × 2) = 28
Number of oxygen atoms = (2 × 19) = 38
Hence, the number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively
Answer: the empirical formula is C3H4O3
Explanation:Please see attachment for explanation
Answer:
78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene
Explanation:

Moles of propene = 
According to reaction, 1 mole of propene gives 1 mole of propane.
Then 0.6538 moles of bromo-propane will give:

78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.
<span>The energy associated with the 590nm atomic emission line is 3.369 x 10^49 Joules. This can be determined by multiplying Plank’s constant (6.626 x 10^34 Js) by the speed of light (3x 10^8 m/s), and then dividing the result by the wavelength (590 x 10^-9 m).</span>