<u>Given:</u>
Calculated density values-
Aluminum = 2.7 g/cm3
Copper = 9.0 g/cm3
Iron = 7.9 g/cm3
Titanium = 4.8 g/cm3
Unknown sample mass = 9.5 g
Sample volume = 2.1 cm3
<u>To determine:</u>
The identity of the unknown sample
<u>Explanation:</u>
'Density' is a physical parameter which can be used to identify the nature of the unknown substance.
Density = Mass/Volume
For the unknown sample
Density = 9.5 g/2.1 cm3 = 4.52 g/cm3
This matches closely with the calculated density of titanium
Ans: The unknown substance is made of titanium
I believe the answer you are looking for is the 4th one.
Average kinetic energy of a particle :
0.5 mv^2 = kT/2
so the kinetic energy = kT/2
assuming the same value of K
T1 = -49 + 273 = 224
T2 = 287 + 273 = 560
E2 / E1 = kT2 / 2 / kT1 / 2
E2 / E1 = T2 / T1
E2 / E1 = 560 / 224 = 2.5
so the average kinetic energy of the particle increases by 2.5
Answer:
<em> 14, 508J/K</em>
ΔHrxn =q/n
where q = heat absorbed and n = moles
Explanation:
<em>m = mass of substance (g) = 0.1184g</em>
1 mole of Mg - 24g
<em>n</em> moles - 0.1184g
<em>n = 0.0049 moles.</em>
Also, q = m × c × ΔT
<em> Heat Capacity, C of MgCl2 = 71.09 J/(mol K)</em>
<em>∴ specific heat c of MgCL2 = 71.09/0.0049 (from the formula c = C/n)</em>
<em>= 14, 508 J/K/kg</em>
ΔT= (final - initial) temp = 38.3 - 27.2
= 11.1 °C.
mass of MgCl2 = 95.211 × 0.1184 = 11.27
⇒ q = 11.27g × 11.1 °C × <em>14, 508 j/K/kg </em>
<em>= 1,7117.7472 J °C-1 g-1</em>
<em />
<em>∴ ΔHrxn = q/n</em>
<em>=1,7117.7472 ÷ 0.1184 </em>
<em>= 14, 508J/K</em>