Answer:
The limiting reactant is H₂
Explanation:
The reaction of hydrogen (H₂) and carbon monoxide (CO) to produce methanol (CH₃OH) is the following:
2H₂(g) + CO(g) → CH₃OH(g)
From the balanced chemical equation, we can see that 1 mol of CO reacts wIth 2 moles of H₂. So, the stoichiometric ratio is:
2 mol H₂/1 mol CO = 2.0
We have 500 mol of CO and 750 mol of H₂, so we calculate the ratio to establish a comparison:
750 mol H₂/500 mol CO = 1.5
Since 2.0 > 1.5, we have fewer moles of H₂ than are needed to completely react with 500 moles of CO. In fact, we need 1000 moles of H₂ and we have 750 moles. So, the limiting reactant is H₂.
High energy hope this helps
Answer:
The correct answer is "obligatory water reabsorption in the proximal convoluted tubule".
Explanation:
The mechanism for producing concentrated urine cannot include the obligatory reabsorption of water in the proximal convoluted tubule since this process is part of the nephron, the system that filters the blood. Glucose and amino acids are reabsorbed almost entirely, as are approximately 70% of filtered potassium and 80% of bicarbonate.
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Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
Answer: Oxidation number of chlorine in potassium chlorate...
so, oxidation state of chlorine in potassium chlorate is +1. and yea!!
Explanation: hope this help