First let's look at the photosynthesis equation:
6CO₂ + 6H₂O --> C₆H₁₂O₆ + 6O₂
We know that:
CO₂ is carbon dioxide
H₂O is water
<span>C₆H₁₂O₆ is glucose
</span><span>O₂ is oxygen
</span>
We also know that the left side of the reaction is the reactants (which gets used up), and the right side is what is produced.
Now, we can look at each of the choices to see which one matches the equation:
We can eliminate the last two choices automatically because there is no hydrogen atoms present on either side.
The second one is not correct because oxygen is a product, not a reactant.
Therefore, the answer is the first one because carbon dioxide is a reactant, and oxygen is a product.
Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:
n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Answer:
∆H = negative and ∆S = positive.
Explanation:
The reaction given in the question is spontaneous at room temperature ,
hence ,
The the gibbs free energy , i.e. ,∆G will be negative for spontaneous reaction
According to the formula ,
∆G = ∆H -T∆S
The value of ∆G can be negative , if ∆H has a negative value and ∆S has a positive value , because , T∆S , has a negative sign .
Hence , the answer will be , ∆H = negative and ∆S = positive.
Answer:
0.056moles HF and 0.70M
Explanation:
When a strong acid is added to a buffer, the acid reacts with the conjugate base.
In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:
NaF + HCl → HF + NaCl
Initial moles of NaF and HF in 60.0mL of solution are:
NaF:
0.0600L × (0.80mol / L)= 0.048 moles NaF
HF:
0.0600L × (0.80mol / L)= 0.048 moles HF
Then, the added moles of HCl are:
0.0200L × (0.40mol / L) = 0.008 moles HCl.
Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:
<em>0.056moles HF</em>
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In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:
0.056mol HF / 0.0800L = <em>0.70M</em>
