Ammonium chloride is the correct name for NH4CL
Answer:
the normality of the given solution is 0.0755 N
Explanation:
The computation of the normality of the given solution is shown below:
Here we have to realize the two sodiums ions per carbonate ion i.e.
N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)
= 0.1886eq ÷ 0.2500L
= 0.0755 N
Hence, the normality of the given solution is 0.0755 N
Answer:
125.5 ×10^-3 m^3= 0.1255 m^3
Explanation:
Volume=5.6mol×22.414dm^3
=125.5dm^3
Question is incomplete, complete question is;
A 34.8 mL solution of 
(aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of 
? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of (aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
0.044 M is the molarity of (aq).
Answer:
P₅O₁₂
<em>Explanation: </em>
Assume that you have 100 g of the compound.
Then you have 44.7 g P and 55.3 g O.
1. Calculate the <em>moles</em> of each atom
Moles of P = 44.7 × 1/30.97 = 1.443 mol Al
Moles of O = 55.3 × 1/16.00 = 3.456 mol O
2. Calculate the <em>molar ratios</em>.
P: 1.443/1.443 = 1
O: 3.456/1.443 = 2.395
3. Multiply by a number to make the ratio close to an integer
P: 5 × 1 = 5
O: 5 × 2.395 = 11.97
3. Determine the <em>empirical formula
</em>
Round off all numbers to the closest integer.
P: 5
O: 12
The empirical formula is <em>P₅O₁₂</em>.
