Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives 
= 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity = 
= 
= 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
equal to 1
Explanation:
All conversion factors used in any calculation is usually a ratio that is equal to 1. This is because when 1 is used to multiply or divide any number, the number stays the same.
- Conversion factors are useful in making the simplification of an expression very simple. They are usually ratios of 1.
- It follows that a conversion factor is a number that when multiplied or divided by itself will stay the same.
Learn more:
conversion factor: brainly.com/question/555814
#learnwithBrainly
First, we need to get n1 (no.of moles of water ): when
mass of water = 0.0203 g and the volume = 1.39 L
∴ n1 = mass / molar mass of water
= 0.0203g / 18 g/mol
= 0.00113 moles
then we need to get n2 (no of moles of water) after the mass has changed:
when the mass of water = 0.146 g
n2 = mass / molar mass
= 0.146g / 18 g/ mol
= 0.008 moles
so by using the ideal gas formula and when the volume is not changed:
So, P1/n1 = P2/n2
when we have P1 = 1.02 atm
and n1= 0.00113 moles
and n2 = 0.008 moles
so we solve for P2 and get the pressure
∴P2 = P1*n2 / n1
=1.02 atm *0.008 moles / 0.00113 moles
= 7.22 atm
∴the new pressure will be 7.22 atm
Explanation:
According to the Henderson-Hasselbalch equation, the relation between pH and 
is as follows.
pH = 
where, pH = 7.4 and = 7.21
As here, we can use the nearest to the desired pH.
So, 7.4 = 7.21 + 
0.19 =
= 1.55
1 mM phosphate buffer means ![[HPO_{4}]](https://tex.z-dn.net/?f=%5BHPO_%7B4%7D%5D)
+ ![[H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
Therefore, the two equations will be as follows.
= 1.55 ............. (1)
+ = 1 mM ........... (2)
Now, putting the value of from equation (1) into equation (2) as follows.
1.55![[H_{2}PO_{4}] + [tex][H_{2}PO_{4}]](https://tex.z-dn.net/?f=%5BH_%7B2%7DPO_%7B4%7D%5D%20%2B%20%5Btex%5D%5BH_%7B2%7DPO_%7B4%7D%5D)
= 1 mM
2.55 = 1 mM
= 0.392 mM
Putting the value of in equation (1) we get the following.
0.392 mM + = 1 mM
= (1 - 0.392) mM
= 0.608 mM
Thus, we can conclude that concentration of the acid must be 0.608 mM.
