Answer:
Option D.
Explanation:
First we convert the given reactant masses into moles, using their respective molar masses:
- 4.00 g H₂ ÷ 2 g/mol = 2 mol H₂
- 6.20 g P₄ ÷ 124 g/mol = 0.05 mol P₄
0.05 moles of P₄ would react completely with (6*0.05) 0.3 moles of H₂. There are more H₂ moles than required, meaning H₂ is in excess and P₄ is the limiting reactant.
Now we<u> calculate how many PH₃ moles could be formed</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.05 mol P₄ *
= 0.2 mol PH₃
Finally we <u>convert 0.2 mol PH₃ into grams</u>, using its <em>molar mass</em>:
- 0.2 mol PH₃ * 34 g/mol = 6.8 g
So the correct answer is option D.
E is Bohrs model the dots on the rings represent the valence electrons
To calculate this, we need the Molarity formula. This formula tell us that Molarity, which is a concentration unit, is equal to the number of moles divided by the volume. In this question we already have the Molarity and the Volume, so let's build our equation:
C = n/V (You can see Molarity with the letter "C" because it means concentration)
3 = n/1
n = 1 * 3
n = 3 moles of NaOH
The answer is: the mass of 6.02 x 1023 representative particles of the element.
The base SI unit for molar mass is kg/mol, but chemist more use g/mol (gram per mole).
For example, molar mas of ammonia is 17.031 g/mol.
M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.
M(NH₃) = 14.007 + 3 · 1.008 · g/mol.
M(NH₃) = 17.031 g/mol.
The molar mass (M) is the mass of a given substance (in this example ammonia) divided by the amount of substance.
Explanation:
It is known that for high concentration of 
, reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.
Now, 
= 0 and the general reaction equation is as follows.
3.00 M n = 2 30 mM
E = 
= 
= 0.038 V
Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.
