<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
From the Graham's law of effusion;
R1/R2 = √MM2/√MM1
Molar mass of chlorine gas is 71
Therefore;
1.87= √ 71 /√mm1
= 1.87² = 71/mm1
mm1 = 71/1.87²
= 71/3.4969
= 20.3
Thus, the molar mass of the other gas is 20.3 , and i think the gas is neon
Answer:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
Explanation:
The temperature of an object can be written using different temperature scales.
The two most important scales are:
- Celsius scale: the Celsius degree is indicated with ºC. It is based on the freezing point of water (placed at 0ºC) and the boiling point of water (100ºC).
- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.
The expression to convert from Celsius degrees to Kelvin is:
Therefore in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:
1) The Kelvin temperature cannot be negative
2) The Kelvin degree is written as K, not ºK
3Zn + 8HNO3 ---> 3Zn(NO3)2 + 4H2O + 2NO IF IT IS COLD AND DILUT NITRIC ACID .
IF IT IS HOT AND CONCENTRATED THEN:
Zn+ 4HNO3 ---> Zn(NO3)2 +2H2O +2NO2
