Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %
substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span>
Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span>
<span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span>
<span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>
Answer:
explain on which topic we have to write and organize an essay ?
Explanation:
Each of the prefixes means a different thind:
Nano means 0.000000001
Kilo means 1000
Milli means 0.001
Centi means 0.01
Out of these four, the biggest is b: 1 kilogram
Answer:
(i) ΔU = 116 J
(ii) ΔU = 289 J
(iii) ΔU = 1 KJ
(iv) ΔU = 0 J
(v) ΔU = 3.25 KJ
Explanation:
first law:
(i) W = 153 J; Q = - 37 J ( Q ( - ), losing friction )
⇒ ΔU = 153 - 37 = 116 J
(ii) W = 289 J; Q = 0 ( insulated)
⇒ ΔU = W = 289 J
(iii) Q = 1 KJ , W = 0 ( isovolumetric process)
⇒ ΔU = Q = 1 KJ
(iv) isothermal ( constant temperature )
∴ ΔT = 0° ( isothermal )
⇒ ΔU = 0 J
(v) isobaric ( constant pressure )
⇒ ΔU = Q + W
∴ Q = 15.6 KJ
∴ W = - ∫ P dV = - P ΔV; W (-) the system performs a job and the volume increases
.
∴ P = 950 KPa * ( 1000 Pa / KPa ) = 950000 Pa = 950000 J/m³
∴ ΔV = 18 - 5 = 13 L * ( m³ / 1000 L ) = 0.013 m³
⇒ W = - ( 950000 J/m³) * ( 0.013 m³ ) = - 12350 J ( - 12.35 KJ )
⇒ ΔU = 15.6 KJ + ( - 12.35 KJ )
⇒ ΔU = 3.25 KJ
