Answer:
531.6g
Explanation:
Total moles of glucose in this case is: 886/180= 4.922 (mole)
For every 1 mole glucose we get 6 mole water
-> Mole of water is: 4.922 * 6= 29.533 (mole)
weight of water is 18. Therefore, total weight of water that we will have from 886g of glucose are: 25.933*18= 531.6g
The stoichiometric amount of water produced by reacting 886g of glucose in the given reaction is approximately 532g.
This is a question of stoichiometry, the part of chemistry that deals with the relationships between reactants and products in chemical reactions. In this reaction, 1 molecule of glucose (C6H12O6) produces 6 molecules of water (H2O). Looking at the molar mass of glucose, which is approximately 180.16 g/mol, and of water, which is about 18.015 g/mol, we can determine the produced water mass. 886 grams of glucose is approximately 4.92 moles. Because the reaction produces 6 moles of water for every mole of glucose, the reaction of 4.92 moles of glucose will produce approximately 29.52 moles of water. Turning that back into grams, we find that 29.52 moles of water is approximately 532 grams of water. So, the reaction of 886 g of glucose would produce about 532 g of water.
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Answer:
i cant see the picture sorry
Explanation:
b. They contain a carbonyl group with a nonpolar carbon-oxygen bond.
c. The functional group of this type of compound must always be on the end of a carbon chain.
d. The functional group of this type of compound must always be in the middle of a carbon chain.
Answer:
Option d.
Explanation:
Ketones contain a carbonyl groups as a functional group, which is a carbon bonded to oxygen with a double bond. In a ketone, the carbon is always bonded to two carbon atoms:
R-C(=O)-R'
The carbon in the carbonyl group has a hybridization sp2 (3 hybrid orbitals with an unhybridized p orbital), where two of the orbitals form sigma (σ) bonds with the other two carbons (R-C-R') and the other hybrid orbital form a sigma bond with the oxygen (C-O). The unhybridized p orbital on the carbon atom is used to form a pi (π) bond with the oxygen, thus forming the double bond (C=O).
The bond of a carbonyl group is polar, because of the difference of the electronegativity between the carbon atom and the oxygen atom.
Hence, from all of the above we can discard the option a, (the carbonyl groups exhibits sp2 hybridization), the option b (carbon-oxygen bond is a bond polar) and the option c (the group must always be in the middle of a carbon chain, the groups that are always in the end, are a aldehyde groups).
Therefore, the correct option is d, the functional group of this type of compound must always be in the middle of a carbon chain.
I hope it helps you!
Answer:
d. The functional group of this type of compound must always be in the middle of a carbon chain.
Explanation:
1. oak
2. catrpiller
3. Blue jay
Answer:
Because the fungus is killing the oak leaves, the caterpillars will have less food and die out, resulting in a decrease in blue jay populations.
Hope it helps :) and let me know if you want me to elaborate.
Answer:
The final pressure is 0.788 atm (option b).
Explanation:
Boyle's law says that the volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure. That is: if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases. This is expressed mathematically as the product of pressure times volume equal to a constant value:
P*V=k
Assuming a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment, by varying the volume of gas to a new value V2, then the pressure will change to P2, and it will be fulfilled:
P1*V1=P2*V2
In this case:
Replacing:
2.14 atm*3 L= P2* 8.15 L
Solving:
0.788 atm= P2
The final pressure is 0.788 atm (option b).
Answer:
1.7 × 10⁴ J
Explanation:
Step 1: Calculate the heat required to raise the temperature of ice from -15 °C to 0°C
We will use the following expression.
Q₁ = c(ice) × m × ΔT
Q₁ = 2.03 J/g.°C × 25 g × [0°C - (-15°C)] = 7.6 × 10² J
Step 2: Calculate the heat required to melt 25 g of ice
We will use the following expression.
Q₂ = C(fusion) × m
Q₂ = 80. cal/g × 25 g × 4.184 J/1 cal = 8.4 × 10³ J
Step 3: Calculate the heat required to raise the temperature of water from 0°C to 75 °C
We will use the following expression.
Q₃ = c(water) × m × ΔT
Q₃ = 4.184 J/g.°C × 25 g × (75°C - 0°C) = 7.8 × 10³ J
Step 4: Calculate the total heat required
Q = Q₁ + Q₂ + Q₃
Q = 7.6 × 10² J + 8.4 × 10³ J + 7.8 × 10³ J = 1.7 × 10⁴ J
Explanation:
a) Corrosive and b) Reacts with certain metals