Answer:
The final temperature = 848.7 K or 575.7 °C
Explanation:
Step 1: Data given
Temperature = 25.0 °C = 298 K
Volume = 158 mL = 0.158 L
The volume increases to 450 mL = 0.450 L
The pressure is constant
Step 2: Calculate the final temperature
V1/T1 = V2/T2
⇒with V1 = the initial volume = 0.158 L
⇒with T1 = the initial temperature = 298 K
⇒with V2 = the increased volume = 0.450 L
⇒with T2 = the new temperature = TO BE DETERMINED
0.158 L / 298 K = 0.450 L / T2
T2 = 0.450 L / (0.158L/298K)
T2 = 848.7 T
The final temperature = 848.7 K or 575.7 °C
The catabolism of glucose has an equation of C6H12O6 + 6O2 = 6CO2 +6 H20. Hence for every mole of glucose, 6 moles of CO2 is produced. Given 22 grams of CO2, that is 0.5 mol CO2, we multiply this by 1/6, we get the number of moles of glucose equal to 1/12 mol. The mass of glucose needed is obtained by multiplying this by molar mass of glucose which is 180 g/mol. This is equivalent 15 grams of glucose.
For getting the result of this problem, the knowledge of periodic table is very important. From the periodic table we come to know that. The knowledge of atomic mass of magnesium is also required to solve the problem.
1 mole of magnesium = 24.3 gm
88.1 moles of magnesium = (24.3 * 88.1) gms
= 2140.83 gms
So 2140.83 grams are there in 88.1 moles of magnesium.
<h3>
Answer:</h3>
56.11 g/mol
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Compound] KOH
<u>Step 2: Identify</u>
[PT] Molar Mass of K - 39.10 g/mol
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mass of H - 1.01 g/mol
<u>Step 3: Find</u>
39.10 + 16.00 + 1.01 = 56.11 g/mol