Answer:
0.0970 M
Explanation:
Remember this equation:
mol/M x V
Convert it so that you can get M.
M=mol/V
Convert the 2.14 grams of H2SO4 into mols
=0.0218
Convert mL to L
225/1000
=0.225
Plug it in.
0.0218/0.225
=0.0970 M
Answer:
I) the heat capacity of ammonia(s)
II) the heat capacity of ammonia(ℓ)
IV) the enthalpy of fusion of ammonia
Explanation:
Initially, ammonia at 200 K is liquid. To calculate the change of enthalpy from 200 K to 195 K (melting point) we need to know the heat capacity of ammonia(ℓ).
At 195, ammonia is in the transition from liquid to solid (solidification). To calculate the change of enthalpy in that process we need to know the enthalpy of solidification of ammonia, which has the same value but opposite sign to the enthalpy of fusion of ammonia.
From 195 K to 0 K, ammonia is solid. To calculate the change of enthalpy in that process we need to know the heat capacity of ammonia(s).
<u>Answer:</u> The balanced chemical equation is 
<u>Explanation:</u>
A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.
For the given unbalanced chemical equation:

<u>On the reactant side:</u>
Atoms of K = 1
Atoms of Cl = 1
Atoms of O = 3
<u>On the product side:</u>
Atoms of K = 1
Atoms of Cl = 1
Atoms of O = 2
To balance the equation, we must balance the atoms by adding 2 infront of both
and
. Also, a coefficient of 3 must be written infront of 
For the balanced chemical equation:

Answer:
191.11 grams of oxygen gas should be produced.
Explanation:
The balanced reaction is:
2 Al₂O₃ → 4 Al + 3 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Al₂O₃: 2 moles
- Al: 4 moles
- O₂: 3 moles
Being the molar mass of each compound:
- Al₂O₃: 102 g/mole
- Al: 27 g/mole
- O₂: 32 g/mole
By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Al₂O₃: 2 moles* 102 g/mole= 204 grams
- Al: 4 moles* 27 g/mole= 108 grams
- O₂: 3 moles* 32 g/mole= 96 grams
Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

mass of oxygen= 191.11 grams
<u><em>191.11 grams of oxygen gas should be produced.</em></u>