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:
Stars form from an accumulation of gas and dust, which collapses due to gravity and starts to form stars. The process of star formation takes around a million years from the time the initial gas cloud starts to collapse until the star is created and shines like the Sun.
Answer is: mass of silver chloride is 34,82 g.
Chemical
reaction: AgNO₃ +
NaCl → AgCl + NaNO₃.
<span>m(AgNO</span>₃) = 53,42 g..<span>
m(NaCl) = 14,19 g.
n(AgNO</span>₃) = m(AgNO₃) ÷ M(AgNO₃).<span>
n(AgNO</span>₃) = 53,42 g ÷ 169,87 g/mol.
n(AgNO₃) = 0,314 mol.
n(NaCl) = 14,19 g ÷ 58,4 g/mol.
n(NaCl) = 0,242 mol; limiting reactant.
From chemical reaction: n(NaCl) : n(AgCl) = 1 : 1.<span>
n</span>(AgCl)<span> = 0,242 mol.
m</span>(AgCl) = 0,252 mol · 143,32 g/mol.
m(AgCl) = 34,82 g.
Answer:
I don't know if option e. is 22,5 g. because that is the correct answer for this. 15% NaCl (mm) gives you the information that in 100 g of solution, you have 15 g of solute. So, if in 100 g of solution you have 15 g of NaCl, in 150 g of solution, how much. Try the rule of three.
Explanation:
Answer:
0.50 mol
Explanation:
The half-life is <em>the time required for the amount of a radioactive isotope to decay to half that amount</em>.
Initially, there are 8.0 moles.
- After 1 half-life, there remain 1/2 × 8.0 mol = 4.0 mol.
- After 2 half-lives, there remain 1/2 × 4.0 mol = 2.0 mol.
- After 3 half-lives, there remain 1/2 × 2.0 mol = 1.0 mol.
- After 4 half-lives, there remain 1/2 × 1.0 mol = 0.50 mol.
