Answer: the name of the reaction is Combustion
Explanation: Combustion is a reaction in which a substance burns in air (oxygen) to produce CO2 and H20
Answer:
Internal energy
Explanation:
Particles have two main energies that scientists acknowledge.
- The energy that is associated with their movement which we call kinetic energy. It is worth noting that particles have different kinetic energy, however the average kinetic energy that they have is directly proportional to temperature.
- The energy that is due to the interactive forces (attraction or repulsion) between particles is called potential energy
When these two types of energies are considered collectively, we call that internal energy
Answer:
Explanation:
22 Directions: Select the correct food web. In Ottawalley National Park, grass is abundant. Gazelles spend most of the day feeding on grass, their main source of food. Gazelles are hunted and eaten by the park's wild dogs. Lions are a predator of both wild dogs and gazelles. Which of the following food webs best represents the relationships described above? lion lion wild dog - wild dog gazelle gazelle grass grass lion lion wild dog wild dog gazelle gazelle grass grass
Explanation:
By Graham's Law,
We have r1/r2 = √(M2/M1), where M represents the molar mass of the gas and r is the effusion rate.
Since temperature is not a variable here, the sentence is true.
Problem 1.
1) Grams to mole, for this we need to know molar mass of a substance.
<span>45.7 grams of calcium chloride
</span>calcium chloride is CaCl2
Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
45.7 gram* 1 mol/111.1 g =(45.7/111.1) mol CaCl2
2) reaction
3CaCl2 +Al2O3 -----> 3CaO +2AlCl3
from reaction 3 mol 2 mol
from the problem 45.7/111.1 mol x mol
We have a proportion, so
x= (45.7/111.1)mol*2/3 mol AlCl3
3) Find mass of (45.7/111.1)*2/3 mol AlCl3
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol
((45.7/111.1)*2/3 )*133.5g/1mol= 36.6 g AlCl3
Problem 2.
This problem can be solved the way like the first one.
But both problems can be solved different way.
Second way to solve these problems.
M(Na)=23.0 g/mol,
M(Na2O)= 2*23.0 +16.0= 62.0 g/mol
4Na + O2 -----> 2Na2O
moles from reaction 4mol 2 mol
masses from reaction 4mol*23.0g/1mol 2mol*62.0 g/mol
92 g 124 g
masses from the problem 25.6 g x g
Now, we can write a proportion:
92 g Na produce 124 g Na2O
25.6 g Na produce x g Na2O
92/25.6 = 124/x
x=25.6*124/92= 35.5 g Na2O