M = 22.1 g
V = 52.3 mL
D = ?
D = m/V
= 22.1/52.3
= 22.1*10/52.3*10
= 221/523
= 0.4
There. I’m sorry i forgot what exactly was the S.I. unit of density :(
<span>0.0165 m
The balanced equation for the reaction is
AgNO3 + MgCl2 ==> AgCl + Mg(NO3)2
So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights.
Atomic weight silver = 107.8682
Atomic weight chlorine = 35.453
Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol
Now how many moles were produced?
0.1183 g / 143.3212 g/mol = 0.000825419 mol
So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division.
0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m
Rounding to 3 significant figures gives 0.0165 m</span>
Answer:
27.64 liters
Explanation:
From the balanced equation, 2 moles of K2Cr2O7 requires 3 moles of CH3OH.
Mole of CH3OH = 1.9/32.04 = 0.0593 mole
Mole of K2Cr2O7 that will require 0.0593 mole of CH3OH:
2 x 0.0593/3 = 0.0395 mole
mole = molarity x volume
Volume of K2Cr2O7 needed = 0.0395/0.00143
= 27.64 Liter
<em>Hence, 27.64 liters of 0.00143 M K2Cr2O7 will be required to titrate 1.90 g of CH3OH dissolved in 50.0 mL of solution</em>
Mass × velocity = momentum
74 x 15 = momentum
