This temperature in units of degree Celsius is..
-78° Celsius
        
             
        
        
        
A. The longest carbon chain is eight, and it has two methyl groups attached to carbon three, and a special group attached to carbon five. Its two names could be:
3-dimethyl-5-(1-methylethyl)octane
3-dimethyl-5-isopropyloctane
Both of these are correct. This is an alkane, because it has all single bonds.
B. This has a triple bond contained between carbons 2 and 3, and has a methyl group off carbon 4. The longest chain is 5. It’s name is:
4-methyl-2-pentyne
This is an alkene, because of the double bond.
C. This has a double bond contained between carbons 2 and 3, and has a methyl off of four and an methyl off of six. The longest chain is eight (follow the longest chain of carbons).
4,6-dimethyl-2-octene 
This is an alkene, because of the double bond.
D. This has an ethyl group at 1 and a methyl group at 2 (rotate the compound to make it as clean as possible, in this case, the ring is flipped and rotated to make it alphabetical with the smallest numbers possible). The two names are:
1-ethyl-2-methylbenzene
ortho-ethylmethylbenzene
Both are correct, the ortho prefix telling the location of the ethyl and methyl groups. This is an aromatic structure because of its double bonded ring.
E. The longest chain is nine, and has methyls at three, five, and seven, along with a propyl at five. The name is:
3,5,7-trimethyl-5-propylnonane
This is an alkane, due to the single bonds.
Hope this helps!
        
             
        
        
        
Answer:
hola soy jess, tu respuesta esta aqui
¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2
2 mol Ba(OH)₂ × \frac{1molCO_{2} }{1molBa (OH)_{2}} 
1molBa(OH) 
2
 
 
1molCO 
2
 
 = 2 moles CO₂
Explanation:
espero que pueda ayudarte 
hermana/hermano
lo que
hahahaha
 
        
             
        
        
        
When The balanced equation is:
2Al + 3CuCl2 ⇒3 Cu + 2AlCl3
So, we want to find the limiting reactant:
1- no. of moles of 2Al = MV/n = (Wt * V )/ (M.Wt*n*V) = Wt / (M.Wt *n)
        
where M= molarity, V= volume per liter and n = number of moles in the balanced equation.
by substitute: 
∴ no. of moles of 2Al = 0.2 / (26.98 * 2)= 0.003706 moles.
2- no.of moles of 3CuCl2= M*v / n = (0.5*(15/1000)) / 3= 0.0025 moles.
So, CuCl2 is determining the no.of moles of the products.
∴The no. of moles of 3Cu = 0.0025 moles.
∴The no.of moles of Cu= 3*0.0025=  0.0075 moles.
and ∵ amount of weight (g)= no.of moles * M.Wt = 0.0075 * M.wt of Cu
 = 0.0075 * 63.546 =0.477 g