Answer:
<h3>The answer is 113.1 cm³</h3>
Explanation:
The volume of a sphere is given by
![v = \frac{4}{3} \pi {r}^{3} \\](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%7Br%7D%5E%7B3%7D%20%20%5C%5C%20)
where
r is the radius of the sphere
From the question
r = 3 cm
We have
![v = \frac{4}{3} \times {3}^{3} \times \pi \\ = \frac{4}{3} \times 27\pi \\ = 4 \times 9\pi = 36\pi \\ = 113.0973355...](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%20%5Ctimes%20%20%7B3%7D%5E%7B3%7D%20%20%5Ctimes%20%5Cpi%20%5C%5C%20%20%3D%20%20%5Cfrac%7B4%7D%7B3%7D%20%20%5Ctimes%2027%5Cpi%20%20%20%5C%5C%20%20%3D%204%20%5Ctimes%209%5Cpi%20%3D%2036%5Cpi%20%5C%5C%20%20%3D%20113.0973355...)
We have the final answer as
<h3>113.1 cm³</h3>
Hope this helps you
Answer:
Explanation:
Heat of vaporization is the amount of heat required to convert 1 mole of liquid substance to its vapor form.
Given :
Amount of heat required for 1 mol of water = 40.7 kJ/mol
According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number
of particles.
1 mole of
weighs = 18.02 g
Thus we can say:
18.02 g of
requires heat = 40.7 kJ
Thus 5.6 g of
requires heat =
(1kJ=1000J)
Thus the energy required to completely vaporize 5.6 g of water at 100.°C is
So to get the answer use the equation: moles = mass/ RAM the RAM is the relative atomic mass and can be calculated by using the bottom number in the box of the element on the periodic table so for this question the mass of P4 is 161g and the RAM of P4 is 124 (31 X 4) so 161/124 = 1.3 and because you need to work out P2O5 and not P4 2 is half of 4 so you half the moles so the answer should be 0.65