Convert 57.6 L to dm3 and divide it by 24
Here's my best guess
the volume of the unit cell is (385*10^-12)^3=5.7066*10^-29 m^3
multiply by density to get mass
mass = (7 g/cm^3)*(100^3 cm^3 / 1^3 m^3) * 5.7066*10^-29 m^3= 3.99466*10^-22 g
covert to moles
3.99466*10^-22 g * 1 mol / 239.82 g = 1.6657 *10^-24 mol
convert to number of units
1.6657 *10^-24 mol * 6.23*10^23 units/mol = 1.04
385 pm = 3.85*10^(-8) cm
The volume of the unit cell is the cube of that, which is 5.71*10^(-23) cm^3. Since the ratio of mass to volume (i.e. the density) must be the same no matter what amount of TlCl you have, you can say:
7 = x/(5.71*10^(-23)), where x is the mass of the unit cell. Solving for x, you get 4*10^(-22) g.
The mass of a molecule of TlCl is 240 amu, which in grams is 4*10^(-22) g. The mass of the unit cell and the mass of a molecule of TlCl is the same. Therefore there is one formula unit of TlCl per unit cell.
Answer:
1.85 × 10⁻⁶
Explanation:
0.0003 ÷ 162 = 1.851851852 × 10⁻⁶ ⇒ 1.85 × 10⁻⁶
Hope that helps.
Answer:
1.30mL
Explanation:
The equation for the reaction is given below:
H2SO4 + 2KOH —> K2SO4 + 2H2O
From the equation above we obtained the following:
Mole of acid (nA) = 1
Mole of base (nB) = 2
The following data were obtained from the question:
Mb = 0.00945M
Vb =?
Va = 50mL
Ma = 1.23 × 10^−4M
Using MaVa / MbVb = nA/nB, we can calculate the volume of KOH as illustrated below:
MaVa / MbVb = nA/nB
(1.23 × 10^−4 x 50)/0.00945xVb = 1/2
Cross multiply to express in linear form
1.23 × 10^−4 x 50 x 2 = 0.00945xVb
Divide both side by 0.00945
Vb = (1.23 × 10^−4 x 50 x 2) /0.00945
Vb = 1.30mL
