The person who is behind the boat will see much powerful and bigger waves than the person who is in front of the boat. This phenomenon is caused by the Doppler effect which describes the changes in the observed frequency of a wave whenever there is relative motion between the wave source and the observer.
The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
given that :
molarity = 2.16 M
volume = 250 mL = 0.25 L
the molarity is given as :
molarity = number of moles / volumes in L
from this we can calculate the number of moles, we get :
number of moles of K₂Cr₂O₇ = molarity × volume
number of moles of K₂Cr₂O₇ = 2.16 × 0.25
number of moles of K₂Cr₂O₇ = 0.54 mol
Thus, The moles of potassium dichromate , K₂Cr₂O₇ are required to prepare a 250 mL solution of with a concentration of 2.16 M is 0.54 mol.
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Answer:
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Explanation:
Hey there!:
1) The additional stability that accompanies the formation of the network<span>Crystalline is measured as network enthalpy.
</span>2) The reticular energy is the energy released when the solid Crystal isform from separate ions in the gaseous state. Always exothermic.<span>
3) </span>The enthalpy of the network depends directly on the size of the loads and conversely in the distance between the ions .
hope this helps!
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is: