Answer:
b. 7.5 x 10^-3
Explanation:
To solve this problem we need to keep in mind the <em>definition of molarity</em>:
- Molarity = moles of solute / liters of solution
With the above information in mind it is possible to calculate the moles of solute, given the volume (10 mL) and concentration (0.75 M) of the solution:
- First we<u> convert 10 mL to L</u> ⇒ 10 mL / 1000 = 0.01 L
Then we <u>calculate the moles of AgNO₃</u>:
- moles of solute = Molarity * Liters of solution
- 0.01 L * 0.75 M = 7.5x10⁻³ mol AgNO₃
<em>One mole of AgNO₃ contains one mole of Ag⁺</em>, thus the number of Ag⁺ moles is also 7.5x10⁻³.
Answer:
As the police car moves towards them, the pitch of the police car's siren goes up because the sound waves get closer together in front of the police car. When the police car moves away from the person, the pitch goes down because the waves spread out behind the police and that makes the frequency go down.
Explanation:
I think the charge is +2
Hope this helps
Answer: The Lewis structure of Chloroacetate can be found at the attachment below.
Explanation:
CH2ClCOO- The chemical compound is called Chloroacetate.
Reference link for the Chloroacetate structure.
https://www.google.com/search?q=lewis+structure+for+CH2ClCOO-&prmd=ivn&sxsrf=ALeKk03mQcLiY-q5pEriMR0_26ZTXLjmJg:1589680325594&source=lnms&tbm=isch&sa=X&ved=2ahUKEwjfxPPY5LnpAhVloXEKHeAwD-wQ_AUoAXoECA4QAQ&cshid=1589680746615&biw=360&bih=559&dpr=3#
From a stock solution of 3.00 m nitric acid, 9.391 ml of stock solution is needed to create a 0.161 m nitric acid solution, which has a total volume of 175 ml of the diluted solution.
A chemical reagent is present in vast quantities as a stock solution. It has a uniform concentration. Examples of typical stock solutions in laboratories are nitric acid and hydrochloric acid. These play a critical role in creating the titration-related solution preparations.
We know the formula for dilution type problems
M1 VI = M2 V 2
Where,
M, = initial molarity
V , = initial Volume
M2 = final molarity
V 2 = final Volume
Hene given -
M, = 3.00 M
VI = ?
M2 = 0.161M
V 2 = 175 ml
Accordingly ' MI V1 = M2 V 2
V1 =
V1= (0.161M*175ml)/ 3.00M
v1 = 9.391
The required volume of Stock solution is 9.391ml.
Learn more about Stock solution here
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