The most probable answer is: (a) <span>The black pot emits heat at a faster rate than the silver pot. It means the black pot conducts heat at a higher rate than the silver pot, that's why it cooled off faster. It also means that the silver pot retains heat better and is more of an insulator.</span>
Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
Additional information
Relative atomic mass(Ar) : A=7, G=16
The empirical formula : A₂G
<h3>Further explanation</h3>
Given
3.5g of element A
4.0g of element G
Required
the empirical formula for this compound
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
The empirical formula also shows the simplest mole ratio of the constituent elements of the compound
mol of element A :
mol of element G :
mol ratio A : G = 0.5 : 0.25 = 2 : 1
Answer:
3.1 kg
Explanation:
Step 1: Write the balanced combustion equation
C₈H₁₈ + 12.5 O₂ ⇒ 8 CO₂ + 9 H₂O
Step 2: Calculate the moles corresponding to 1.0 kg of C₈H₁₈.
The molar mass of C₈H₁₈ is 114.23 g/mol.
1.0 × 10³ g × 1 mol/114.23 g = 8.8 mol
Step 3: Calculate the moles of CO₂ produced from 8.8 moles of C₈H₁₈
The molar ratio of C₈H₁₈ to CO₂ is 1:8. The moles of CO₂ produced are 8/1 × 8.8 mol = 70 mol.
Step 4: Calculate the mass corresponding to 70 moles of CO₂
The molar mass of CO₂ is 44.01 g/mol.
70 mol × 44.01 g/mol = 3.1 × 10³ g = 3.1 kg
Answer:
I hope this link helps you.
Explanation:
http://astronomy.swin.edu.au/cosmos/P/Phases
