Answer:
Emission of Light
Explanation:
Chemiluminescence is the emission of light as the result of a chemical reaction, and not a property of a specific compound.
Dont know if you were asking for this but hope it helps
A− A) A nonmetal that gained one electron
B+ B) A metal that lost one electron
C2− C) A metal that lost two electrons
D2+ D) A nonmetal that gained two electrons
Answer:
c2-c
Explanation:
A mental that lost two electrons just joking I don't k ow but I think it ain't the first o e because of comment sense to the 5th power
Answer:
A and B
Explanation:
metal form ion by lossing elecron
non metal form ion by gaining electron
Answer:
Crust.
Explanation:
Its is made up hard rocks
Answer:lithosphere
The lithosphere is the solid, outer part of the Earth. The lithosphere includes the brittle upper portion of the mantle and the crust, the outermost layers of Earth's structure.
Explanation:
Answer:
226.2 kJ/mol
Explanation:
Let's consider the following thermochemical equation for the combustion of acetylene.
C₂H₂(g) + (5/2) O₂(g) → 2 CO₂(g) + H₂O(l), ΔH°rxn = –1299 kJ/mol.
We can also calculate the enthalpy of the reaction per mole of acetylene from the enthalpies of formation.
ΔH°rxn = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(C₂H₂(g)) - 1 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₂H₂(g)) = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - ΔH°rxn - 1 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₂H₂(g)) = 2 mol × (-393.5 kJ/mol) + 1 mol × (-285.8 kJ/mol) - (-1299 kJ) - 1 mol × (0 kJ/mol)
ΔH°f(C₂H₂(g)) = 226.2 kJ/mol
Answer:
The enthalpy of formation of acetylene is 226.2 kJ/mol
Explanation:
Step 1: Data given
C2H2 (g) + (5/2)O2 (g) → 2CO2 (g) + H2O (l) Heat of Reaction (Rxn) = -1299kJ/mol
Standard formation [CO2 (g)]= -393.5 kJ/mol
Standard formation [H2O (l)] = -285.8 kj/mol
Step 2: The balanced equation
The formation of acetylene is:
2C(s) + H2(g) → C2H2(g)
Step 3: Calculate the enthalpy of formation of acetylene
It doesn't matter if the process will happen in 1 step or in more steps. What matters is the final result. This is Hess' law of heat summation.
To have the reaction of the formation of acetylene we have to take:
⇒ the reverse equation of the combustion of acetylene
2CO2 (g) + H2O (l) → C2H2 (g) + (5/2)O2 (g)
⇒ The equation of formation of CO2 (multiplied by 2)
2C(s) + 2O2(g) → 2CO2(g)
⇒ the equation of formation of H2O
H2(g) + 1/2 O2(g) → H2O(l)
2CO2 (g) + H2O (l) + 2C(s) + 2O2(g) + H2(g) + 1/2 O2(g → C2H2 (g) + (5/2)O2 (g) + 2CO2(g) + H2O(l)
Final reaction = 2C(s) + H2(g) → C2H2(g)
Calculate the enthalpy of formation of acetylene =
ΔHf = 1299 kJ/mol + (2*-393.5) kJ/mol + (-285.8) kJ/mol
ΔHf = 226.2 kJ/mol
The enthalpy of formation of acetylene is 226.2 kJ/mol
B. CS2 + 3O2 yields CO2 + 2SO2
C. Mg(ClO3)2 yields MgCl2 + 2O2
D. Zn + H2SO4 yields H2 + ZnSO4
There is 65% of NaHCO3 in the sample.
The equation of the reaction is;
HA + NaHCO3 -----> NaA + CO2 + H2O
Amount of CO2 formed = mass/molar mass
mass of CO2 = 0.561 g/44 g/mol = 0.013 moles
From the balanced reaction equation;
1 mole of NaHCO3 yields 1 mole of CO2
0.013 moles of Na2CO3 yields 0.013 moles of CO2
Hence, mass of NaHCO3 in the sample = 0.013 moles × 84 g/mol = 1.092 g of NaHCO3
Percentage by mass of NaHCO3 = 1.092 g/1.68 g ×100/1
= 65%
Learn more: brainly.com/question/25150590
Answer:
63.75%.
Explanation:
The first step here is to write out the reaction showing the chemical reaction between the two chemical species. Thus, we have;
HA(aq) + NaHCO3 --------------> CO2(g) + H20(l) + NaA(aq).
Therefore, the mole ratio is 1 : 1 : 1 : 1 that is go say one mole of HA reacted with one mole of NaHCO3 to give one mole of CO2 and one .ole of NaA.
Hence, the number of moles of CO2 = mass/molar mass = 0.561/44 = 0.01275 moles.
Thus, the number of moles of NaHCO3 = number of moles of CO2 = 0.01275 moles.
Therefore, we have ( 0.01275 moles × 84 g/mol) grams = 1.071 g NaHCO3 in the mixture.
Therefore, the percent by mass of N a H C O 3 in the original mixture = 1.071/1.68 × 100 = 63.75%.