Answer:
synthesis
Explanation:
The reaction given is a synthesis reaction or combination reaction.
Given equation:
NH₃ + HCl → NH₄Cl
In a synthesis reaction, a single product is formed from two or more reactants
A + B → C
The formation of compounds from the union of constituent elements also falls into this category.
So the given reaction is a synthesis reaction.
In a decomposition reaction two or more products are formed from a single reactant
In a single replacement reaction, one substance replaces another.
In double replacement reactions partners in a chemical specie are exchanged.
Therefore the given reaction is a synthesis reaction
Answer:
0.4429 m
Explanation:
Given that mass % of the ethanol in water = 2%
This means that 2 g of ethanol present in 100 g of ethanol solution.
Molality is defined as the moles of the solute present in 1 kilogram of the solvent.
Given that:
Mass of 
= 2 g
Molar mass of = 46.07 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of water = 100 - 2 g = 98 g = 0.098 kg ( 1 g = 0.001 kg )
So, molality is:
<u>Molality = 0.4429 m</u>
The reaction between C2H2 and O2 is as follows:
2C2H2 + 5O2 = 4CO2 + 2H2O
After balancing the equation, the reaction ratio between C2H2 and O2 is 2:5.
The moles of O2 in this reaction is 84.0 mol. According to the above ratio, the moles of C2H2 needed to react completely with the O2 is 84.0mole *2/5 = 33.6 mole.
Answer:
ΔH°f P4O10(s) = - 3115.795 KJ/mol
Explanation:
- P4O10(s) + 6H2O(l) ↔ 4H3PO4(aq)
- ΔH°rxn = ∑νiΔH°fi
∴ ΔH°rxn = - 327.2 KJ
∴ ΔH°f H2O(l) = - 285.84 KJ/mol
∴ ΔH°F H3PO4(aq) = - 1289.5088 KJ/mol
⇒ ΔH°rxn = (4)(- 1289.5088) - (6)(- 285.84) - ΔH°f P4O10(s) = - 327.2 KJ
⇒ ΔH°f P4O10(s) = - 5158.035 + 1715.04 + 327.2
⇒ ΔH°f P4O10(s) = - 3115.795 KJ/mol
Hey there!
Al + HCl → H₂ + AlCl₃
Balance Cl.
1 on the left, 3 on the right. Add a coefficient of 3 in front of HCl.
Al + 3HCl → H₂ + AlCl₃
Balance H.
3 on the left, 2 on the right. We have to start by multiplying everything else by 2.
2Al + 3HCl → 2H₂ + 2AlCl₃
Now we have 2 on the right and 4 on the left. Change the coefficient in front of HCl from 3 to 4.
2Al + 4HCl → 2H₂ + 2AlCl₃
Now, for Cl, we have 4 on the left and 6 on the right. Change the coefficient in front of HCl again from 4 to 6.
2Al + 6HCl → 2H₂ + 2AlCl₃
Now, our H is unbalanced again. 6 on the left, 4 on the right. Change the coefficient in front of H₂ from 2 to 3.
2Al + 6HCl → 3H₂ + 2AlCl₃
Balance Al.
2 on the left, 2 on the right. Already balanced.
Here is our final balanced equation:
2Al + 6HCl → 3H₂ + 2AlCl₃
Hope this helps!
