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:
A beaker
Step-by-step explanation:
Specifically, I would use a 250 mL graduated beaker.
A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.
You don't need precisely 100 mL solution.
If the beaker is graduated, you can easily measure 100 mL of the stock solution.
Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).
Answer:
... chloride, calcium, potassium, and zinc was signifi- ... of cow and goat milk pasteurization on element retention ... certified American Chemical Society (ACS);. Whatman ... goat milk. Table 2 gives the content of 17 elements of ... found .0026 rag/100 g in raw and .0024 mg/100 ... mg/100 g chloride content (27) and another.
Altitude, If this helped please 5 star me
Let suppose the Gas is acting Ideally, Then According to Ideal Gas Equation,
P V = n R T
Solving for P,
P = n R T / V ----- (1)
Data Given;
Moles = n = 1.20 mol
Volume = V = 4 L
Temperature = T = 30 + 273 = 303 K
Gas Constant = R = 0.08206 atm.L.mol⁻¹.K⁻¹
Putting Values in Eq.1,
P = (1.20 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 303 K) ÷ 4 L
P = 7.45 atm
