Answer:
607 ppm
Explanation:
In this case we can start with the <u>ppm formula</u>:

If we have a solution of <u>0.0320 M</u>, we can say that in 1 L we have 0.032 mol of
, because the molarity formula is:

In other words:


If we use the <u>atomic mass</u> of
(19 g/mol) we can convert from mol to g:
Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

Finally we can <u>divide by 1 L</u> to find the ppm:

<u>We will have a concentration of 607 ppm.</u>
I hope it helps!
Energy, mass, and the speed of light squared.
(E, M, and C^2 respectively).
Hope this helped! :)
Answer is B. gas formation
Answer:
B) 0.6M
Explanation:
I apologize in advance if it is not correct :l
The (M1V1= M2V2) is given for you to plug in the correct numbers so let's jot this down.
(M1*V1= M2*V2)
so they give us 6M which would be our (M1), from this we can also conclude that 5mL is also V1; ( if you notice the M1's and V1's are always found next to eachother). This leads us to our 50mL, this would be our V2 because the volume went from 5mL to 50mL. Now lets put this in order based on what we know.
M1= 6M (M1*V1= M2*V2)
V1= 5mL
M2= ?
V2= 50mL
now we plug in what we know into the equation to find the unknown (M2)
(6M*5mL= M2*50mL)
now we could do the long math, but I don't think your on brainly to do the hard way. so lets keep it simple!
We are going to put the 50mL under the (6M*5mL) for division.
This is honestly MUCH easier, than manually answering. you just put that in the calculator and it'll give you B) 0.6M
honestly though I might not know what I'm doing cuz im currently doing my test and decided to answer this question ;)
Good Luck!
Answer:
They both have the same number of atoms
Explanation:
The number that indicates the amount of particles in a compound is the Avogadro's number (NA).
It does not matter the mass of compound we have, If we have 1 mol we will be sure that we are talking about 6.02×10²³ particles
6.02×10²³ represents the amount of atoms in twelve grams of 12-pure carbon and it is considered a reference to measure the amount of all kinds of substances present in a given system.