Answer:
13.5g of AgNO3 will be needed
Explanation:
Silver nitrate, AgNO3 contains 1 mole of silver, Ag, per mole of nitrate. To solve this problem we need to convert the mass of Ag to moles. Thee moles = Moles of AgNO3 we need. With the molar mass of AgNO3 we can find the needed mass:
<em>Moles Ag-Molar mass: 107.8682g/mol-</em>
8.6g * (1mol / 107.8682g) = 0.0797 moles Ag = Moles AgNO3
<em>Mass AgNO3 -Molar mass: 169.87g/mol-</em>
0.0797 moles Ag * (169.87g/mol) =
<h3>13.5g of AgNO3 will be needed</h3>
Answer:
Isotopes of an element share the same number of protons but have different numbers of neutrons. ... This means that all three isotopes have different atomic masses (carbon-14 being the heaviest), but share the same atomic number (Z=6).
Explanation:
Answer:
is this based on the newtons law and balnce force
Explanation:
Answer:
pH = 11.7
Explanation:
<u>Given:</u>
[OH-] = 0.00500 M
<u>To determine:</u>
pH of the solution
<u>Explanation:</u>
pH refers to the H+ ion concentration in a given solution whereas pOH indicates the OH- ion concentration.
pH and pOH are related by the following equation:

Here:
![pH = -log[H+]\\\\pOH = -log[OH-]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%2B%5D%5C%5C%5C%5CpOH%20%3D%20-log%5BOH-%5D)
When OH- = 0.00500 M

Based on equation(1):

Answer : The correct option is, Only Student B
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, 
As we know that nitrogen has '5' valence electrons and hydrogen has '1' valence electron.
Therefore, the total number of valence electrons in
= 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
The Lewis dot structure of student A is wrong because there is a coordinate bond present between the nitrogen and hydrogen is not covalent.
The correct Lewis-dot structure of
is shown by the student B.