The rate of change in the amount of silver supplied is equal to the ratio of the difference in the amount supplied to the number of years elapsed from 2005 to 2008. That is,
rate of change = (25 - 200) / (2008 - 2005) = -173/3
The answer is letter A.
H₂O would be the limiting reactant.
Balanced chemical equation:
6CO₂ + 6H₂O + light equation → C₆H₁₂O₆ + 6O₂
The amount of product that can be created is constrained by the reactant that is consumed first in a chemical reaction, commonly referred to as the limiting reactant (or limiting reagent).
Given
No. of moles of CO₂ = 18.6
Mass of H₂O = 2.38 × 10² g = 238g
No. of moles of H₂O = Given mass/ Molar mass
= 238 / 18 = 13.22 moles
Moles of H₂O = 13.22
According to the balanced chemical equation
6 moles of CO₂ react with 6 moles of H₂O
So the reactant that has less number of moles will be consumed first.
As the No. of moles of H₂O < No. of moles of CO₂
So, H₂O is the limiting reactant with 13.22 moles.
Hence, H₂O would be the limiting reactant.
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Answer: 1. 
2. 3 moles of 
: 2 moles of 
3. 0.33 moles of : 0.92 moles of
4. is the limiting reagent and is the excess reagent.
5. Theoretical yield of 
is 29.3 g
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
3 moles of require = 2 moles of
Thus 0.33 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 3 moles of give = 2 moles of
Thus 0.33 moles of give = of
Theoretical yield of 
Thus 29.3 g of aluminium chloride is formed.
Answer:
0!
Explanation:
- You need to search your pKa values for Asn (2.14, 8.75), Gly (2.35, 9.78) and Leu(2.33, 9.74), the first value corresponding to -COOH, the second to -NH3 (a third value would correspond to an R group, but in this case that does not apply), and we'll build a table to find the charges for your possible dissociated groups at indicated pH (7), we need to remember that having a pKa lower than the pH will give us a negative charge, having a pKa bigger than pH will give us a positive charge:
-COOH -NH3
pH 7------------------------------------------------------
Asn - +
Gly - +
Leu - +
- Now that we have our table we'll sketch our peptide's structure:
<em>HN-Asn-Gly-Leu-COOH</em>
This will allow us to see what groups will be free to react to the pH's value, and which groups are not reacting to pH because are forming the bond between amino acids. In this particular example only -NH group in Ans and -COOH in Leu are exposed to pH, we'll look for these charges in the table and add them to find the net charge:
+1 (HN-Asn)
-1 (Leu-COOH)
=0
The net charge is 0!
I hope you find this information useful and interesting! Good luck!
Answer:
there is two mixture homogeneous and heterogeneous mixture
ok
