Answer:
64.7g
Explanation:
The balanced chemical equation of this question is as follows;
AlI + HgCl2 → HgI + AlCl2
Based on the above equation, 1 mole of AlI (aluminum monoiodide) reacts to produce 1 mole of HgI (mercury iodide).
Using mole = mass/molar mass to convert mass of HgI to moles.
Molar mass of HgI = 200.59 + 127
= 327.59g/mol
Mole = 138/327.59
= 0.42mol
- If 1 mole of AlI (aluminum monoiodide) reacts to produce 1 mole of HgI (mercury iodide)
- Then 0.42 mol of HgI will be produced by 0.42mol of AlI.
Using mole = mass/molar mass
Mass = mole × molar mass
Molar mass of AlI = 27 + 127
= 154g/mol
Mass of AlI = 0.42 × 154
= 64.7g of AlI
Answer:
To determine the correct subscripts in a chemical formula, you have to solve how many atoms you need to balance the charge.
For example if I had the compound Calcium Fluoride I would look at the periodic table and see that Calcium's ionic formula is
Ca2+
. How do I know this? Well all elements want to have 8 valance electrons so they can be stable(happy). Seeing that Calcium has 2 valance electrons it is going to give away 2 electrons because that is easier than gaining 6 to be happy. Since Calcium has given away 2 electrons it has two more protons than electrons. We know that Protons have a Positive charge, Electrons have Negative charge, and the number of electrons is equal to the atomic number of an element in its pure non-ionic state. (Meaning it doesn't have a positive or negative charge; it is balanced.)
So if calcium gave away two electrons, it will have two more protons than an electron giving it a (2+) charge. The same process can be applied to Fluoride. Since fluoride is one to the left of the noble gases(group 18 or 8A) on the periodic table we know that it has 7 valance electrons because it is in group 7A or 17.
Knowing that we have 7 electrons the fluoride atom will gain an extra electron. Since the fluoride atom gained an extra electron it will have one more negative charge than a positive making it a ^(−)ion.
So you know that Calcium has a 2+ charge and that fluoride has a 1- charge, you then need these ions to balance out. So you need two fluorine atoms with a 1- ions to balance out the 2+ ion of calcium. Your final answer would be
CaF2
because you need two fluorine atoms to balance out the 2+ charge of the calcium.
Final Tip: Determine the charges then inverse the charges, remove the positive and negative superscipts, and write the charge numbers as a sub script. Ie. Calcium Fluoride
Ca2+ and F−
inversing and removing the charge signs would give you
Answer:
Explanation:
The substances that are present at the beginning are called reactants and the substances present at the end are called products. Sometimes when reactants are put into a reaction vessel, a reaction will take place to produce products. Reactants are the starting materials, that is, whatever we have as our initial ingredients.Oxygen, a colorless, odorless, tasteless gas essential to living organisms, being taken up by animals, which convert it to carbon dioxide; plants, in turn, utilize carbon dioxide as a source of carbon and return the oxygen to the atmosphere. Oxygen forms compounds by reaction with practically any other element.
Answer:
4600s
Explanation:
For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:
![-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BB%5D%20-%20-%20-%20%20-%5Cfrac%7Bd%5BB%5D%7D%7B%5BB%5D%7D%3Dk%2Adt)
If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.
PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.
![-\frac{d[P(B)]}{P(B)}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28B%29%5D%7D%7BP%28B%29%7D%3Dk%2Adt)
![-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3Dk%2Adt)
Integrating we get:
![\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ep%20%5C%2C-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3D%5Cint%5Climits%5E%20t%20k%2Adt)
![-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})](https://tex.z-dn.net/?f=-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%3Dk%28t_%7B2%7D-t_%7B1%7D%29)
Clearing for t2:
![\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%7D%7Bk%7D%2Bt_%7B1%7D%3Dt_%7B2%7D)
![ln[P(N_{2}O_{5})]=ln(650)=6.4769](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%3Dln%28650%29%3D6.4769)
![ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%5D%3Dln%28760%29%3D6.6333)
