To do this problem it is necessary to take into account that the heat given by the unknown substance is equal to the heat absorbed by the water, but considering the correct sign:
Clearing the specific heat of the unknown substance:
Answer:
113.69°k
Explanation:
V1=85L of helium V2=32L
T1= 29°C +273= 302°K T2=?
T2=<u>TIV2</u>
V1
T2=<u>(302)(32)</u>= <u>9664</u>
85 85
T2= 113.69°K
I think that it might represent Velocity
Just look at the number in front also called coefficient (you have to balance the equations first, but all the questions here are balanced, so no worries). for q1.
in the balanced equation, the number in front of aluminum oxide is 2 (2 - this number Al2O3) and for aluminium is 4 as in (4 Al). so the ratio is 2:4. simplified it is 1:2. or write it out fully
2 Al2O3: 4 Al
ignore everything after the number.
2:4
same as 1:2
Aluminium oxide to oxygen
2 Al2O3: 3 O2
2:3
aluminum to oxygen
4 Al: 3 O2
4:3
question 2
Mercury oxide to Mercury
2 HgO : 2 Hg
2:2
same as 1:1
Mercury oxide to oxygen
2 HgO : O2
since oxygen in this case does not have a number written in front of it, the default is 1.
2: 1.
you should be able to do the rest
Answer:
8.1 × 10² g
Explanation:
Step 1: Write the balanced equation
2 C₅₇H₁₁₀O₆ + 163 O₂ ⇒ 114 CO₂ + 110 H₂O
Step 2: Convert 1.6 lb of C₅₇H₁₁₀O₆ to g
We will use the conversion factor 1 lb = 453.592 g.
1.6 lb × 453.592 g/1 lb = 7.3 × 10² g
Step 3: Calculate the moles corresponding to 7.3 × 10² g of C₅₇H₁₁₀O₆
The molar mass of C₅₇H₁₁₀O₆ is 890.83 g/mol.
7.3 × 10² g × 1 mol/890.83 g = 0.82 mol
Step 4: Calculate the moles of water produced from 0.82 moles of C₅₇H₁₁₀O₆
The molar ratio of C₅₇H₁₁₀O₆ to H₂O is 2:110. The moles of H₂O produced are 110/2 × 0.82 mol = 45 mol
Step 5: Calculate the mass corresponding to 45 moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
45 mol × 18.02 g/mol = 8.1 × 10² g
