Answer:
1.14atm
Explanation:
Given parameters:
V1 = 250cm³ ;
1000cm³ = 1dm³; so this is 0.25dm³
P1 = 760torr
760torr = 1atm
V2 = 220cm³ ; 0.22dm³
Unknown:
New pressure = ?
Solution:
To solve this problem, we apply Boyle's law and we use the expression below:
P1 V1 = P2V2
The unknown is P2;
1 x 0.25 = P2 x 0.22
P2 = 1.14atm
Mole ratio:
MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl
2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)
moles KOH = 4 x 2 / 1
= 8 moles of KOH
molar mass KOH = 56 g/mol
mass of KOH = n x mm
mass of KOH = 8 x 56
= 448 g of KOH
hope this helps!
Answer:
The mass is 1.4701 grams and the moles is 0.01.
Explanation:
Based on the given question, the volume of the solution is 100 ml or 0.1 L and the molarity of the solution is 0.100 M. The moles of the solute (in the given case calcium chloride dihydride (CaCl2. H2O) can be determined by using the formula,
Molarity = moles of solute/volume of solution in liters
Now putting the values we get,
0.100 = moles of solute/0.1000
Moles of solute = 0.100 * 0.1000
= 0.01 moles
The mass of CaCl2.2H2O can be determined by using the formula,
Moles = mass/molar mass
The molar mass of CaCl2.2H2O is 147.01 gram per mole. Now putting the values we get,
0.01 = mass / 147.01
Mass = 147.01 * 0.01
= 1.4701 grams.
Answer: crest I don’t really know how to explain it but yea it’s crest
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
