Answer:
0.677 moles
Explanation:
Take the atomic mass of K = 39.1, O =16.0, P = 31.0
no. of moles = mass / molar mass
no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)
= 0.02256 mol
From the equation, the mole ratio of KOH : K3PO4 = 3 :1,
meaning every 3 moles of KOH used, produces 1 mole of K3PO4.
So, using this ratio, let the no. of moles of KOH required to be y.
y = 0.02256 x3
y = 0.0677 mol
If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.
Answer:
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom. During the fusion process, radiant energy is released.
Explanation:
The filament holds up the anther so that pollination and fertilization can occur!
Potassium or any other metals.
<u>Answer:</u> The weak bond having slightly positive region and a slightly negative region is polar covalent bond.
<u>Explanation:</u>
Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. <u>For Example:</u> HCl, 
etc..
They are of two types:
- <u>Polar covalent bond:</u> This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond.
In this bond, a dipole is created. More electropnegative atom will have a slight negative charge and less electronegative atom will have a slight positive charge. For Example: 
etc..
- <u>Non-polar covalent bond:</u> This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. For Example:
etc..
Hence, the weak bond having slightly positive region and a slightly negative region is polar covalent bond.
