Given :
Volume of NaCl solution 2.5 L .
Molarity of NaCl solution is 0.070 M .
To Find :
How many moles are present in the solution.
Solution :
Let, n be the number of moles.
We know, molarity is given by :

So,

Therefore, number of moles of NaCl is 0.175 moles.
Use the equation d=m/v
your mass or "m" is 78 g
your volume or "v" is 60mL
if you plug those values into the equation it will look like this:
d=78/60
d=1.3g/mL should be what you come up with
Answer:
The answer to your question is 245 grams
Explanation:
Data
Volume 6.5 L
Molarity = 0.34
mass of CaCl₂ = ?
Process
1.- Calculate the molar mass of CaCl₂
molar mass = (1 x 40) + (2 x 35.5)
= 40 + 71
= 111 g
2.- Convert the grams to moles
111 g of CaCl₂ -------------- 1 mol
x ---------------0.34 mol
x = (0.34 x 111) / 1
x = 37.74 g
3.- Calculate the total mass
37.74 g ------------------ 1 L
x ------------------ 6.5 L
x = (6.5 x 37.74) / 1
x = 245.31
Answer:
The number of lines possible for SO2 is 3
Explanation:
The following Procedure should be followed when calculating the number of vibrational modes:-
- Identify if the given molecule is either linear or non-linear
- Calculate the number of atoms present in your molecule
- Place the value of n in the formula and solve.
SO2 is a non-linear molecule because it contains a lone pair which causes the molecule to bent in shape hence, The mathematical formula for calculating the number of non-linear molecule in a infrared region is (3n - 6) here n is the number of atoms in molecule.
hence for Sulphur Dioxide (SO2), n will be 3
<u> Therefore, The number of lines possible for SO2 is (3*3) - 6 = 3</u>
Answer:
Pentan-2-ol
Explanation:
On this reaction, we have a <u>Grignard reagent</u> (ethylmagnesium bromide), therefore we will have the production of a <u>carbanion</u> (step 1). Then this carbanion can <u>attack the least substituted carbon</u> in the epoxide in this case carbon 1 (step 2). In this step, the epoxide is open and a negative charge is generated in the oxygen. The next step, is the <u>treatment with aqueous acid</u>, when we add acid the <u>hydronium ion</u> (
) would be produced, so in the reaction mechanism, we can put the hydronium ion. This ion would be <u>attacked by the negative charge</u> produced in the second step to produce the final molecule: <u>"Pentan-2-ol".</u>
See figure 1
I hope it helps!