Carbon dioxide (CO2)
in the process of respiration, oxygen and glucose react to form carbon dioxide and water.
Answer:
On the outside or something like that
Answer:
8.934 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 192.12 44.01
H₃C₆H₅O₇ + 3NaHCO₃ ⟶ Na₃C₆H₅O₇ + 3H₂O + 3CO₂
m/g: 13.00
For ease of writing, let's write H₃C₆H₅O₇ as H₃Cit.
(a) Calculate the <em>moles of H₃Cit
</em>
n = 13.00 g × (1 mol H₃Cit /192.12 g H₃Cit)
n = 0.067 67 mol H₃Cit
(b) Calculate the <em>moles of CO₂
</em>
The molar ratio is (3 mol CO₂/1 mol H₃Cit)
n = 0.067 67 mol H₃Cit × (3 mol CO₂/1 mol H₃Cit)
n = 0.2030 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
m = 0.2030 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
m = 8.934 g CO₂
Explanation:
(Ques- A) Why does the first method for determining volume work only for a regular-shaped object?
<u>(Ans- A)</u> <em>Because the method requires precise dimensions of objects for result, which is not possible for irregular shaped objects.</em>
(Ques - B) Will the second method for determining volume work for any object or just an odd-shaped one? Why?
<u>(Ans-B)</u> <em>It will work for both regular and irregular shaped objects since both displace equal volumes of water.</em>
(Ques - C) Is one method of measurement more accurate than the other? Why or why not?
<u>(Ans-C)</u> <em>Both are pretty accurate, with some experimental errors which may creep in accidentally. </em>
(Ques- D) Would the displacement method of measurement work for a cube of sugar? What about a cork? Why?
<u>(Ans - D)</u> <em>No, the method would not work because sugar being soluble, will dissolve in water. </em>
<em>No, the method would not work because sugar being soluble, will dissolve in water. Cork is less dense than water so floats on it, with only part of it submerged in water, resulting in displacement of less volume of water than actual volume of Cork.</em>
(Ques-E) What did you find out from this investigation? Be thoughtful in your answer.
<u>(Ans- E)</u> <em>I learnt about determining volume of different objects from this investigation. </em>(Sorry, I know its not a very thoughtful answer)
Answer:
1597,5 gm AlCl3 decomposed
Explanation:
2AlCl3 --> 2Al + 3Al2 If you produce 15 moles of Al, how many grams of AlCl3 decomposed?
FIRST, you need to write down the the correct reaction. Your equation makes no sense
here is the correct question
2AlCl3 --> 2Al + 3Cl2 If you produce 15 moles of Al, how many grams of AlCl3 decomposed?
for every mole of Al produced, we decomposed a mole of AlCl3
so we produced 15 moles of Al, so we decomposed 15 moles of AlCl3
15 moles are 15 X 106.5 = 1597,5 gm AlCl3 decomposed
