Answer:
ΔH = - 2020.57 kJ/mol
Explanation:
Given that :
mass of propanol = 1.685 g
the molar molar mass = 60 g/mol
Thus; the number of moles = mass/molar mass
= 1.685 g/60 g/mol
= 0.028 g/mol
However ;
ΔH = heat capacity C × Δ T
Given that:
The temperature increases from 298.00 K to 302.16 K.
Then ;
Δ T = 302.16 K - 298.00 K
Δ T = 4.16 K
heat capacity C = 13.60 kJ/K
∴
ΔH = 13.60 kJ/K × 4.16 K
ΔH = 56.576 kJ
The equation of the given reaction can be represented as :
Thus for 0.028 mol of heat liberated; ΔH = 56.576 kJ
For 1 mole of heat liberated now:
ΔH = 56.576 kJ/0.028 mol
ΔH = 2020.57 kJ/mol
SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;
ΔH = - 2020.57 kJ/mol
Answer:
The average of given values is 2.1221 ml
Explanation:
Given data:
Given measurements = 3.00 ml , 2.0 ml, 2.987 × 10⁻³ml , 3.4856 ml
Average value of given measurements = ?
Solution:
Formula:
Average value = sum of all measurement / total number of measurements
2.987 × 10⁻³ml = 0.002987 ml
Now we will put the values.
Average value = 3.00 ml + 2.0 ml + 0.002987 ml+ 3.4856 ml / 4
Average value = 8.488587 ml / 4
Average value = 2.1221 ml
The average of given values is 2.1221 ml
Answer:
Explanation:
To convert from grams to atoms, first divide by the molar mass, the multiply by 6.022*10^23.
To convert from moles to mass, multiply by the molar mass of the element.
Hope this helps!
-Emma Victoria
Physical change .. you cannot undo chemical ... physical is like bending , shattering a mirror , denting something ..
chemical would be rust or something .
Answer:
Yes.
Explanation:
It should be noted that the meaning of molarity is the ratio of moles of solute per liter of solution.
It should be understood that when determining or finding the molarity of an unknown compound ,the process should be performed or carried out at least 3 times. This is done to remove any form of doubt.
The first calculated value for the concentration of the compound will be regarded as rough value, while the second and the third will be regarded as the first and second values respectively.
In this case, the third value for the concentration of HCl will be calculated to for confirmation of other value, that is to be finally sure of its concentration.
