Answer:
(C) Mass of KCl(s), mass of H20, initial temperature of the water, and final temperature of the solution
Explanation:
molar enthalpy of solution of KCl(s) is heat evolved or absorbed when one mole of KCl is dissolved in water to make pure solution . The heat evolved or absorbed can be calculated by the following relation.
Q = msΔt where m is mass of solution or water , s is specific heat and Δt is change in temperature of water .
So data required is mass of water or solution , initial and final temperature of solution , specific heat of water is known .
Now to know molar heat , we require mass of solute or KCl dissolved to know heat heat absorbed or evolved by dissolution of one mole of solute .
Answer: C2H4 + 3O2 → 2CO2 + 2H2O
Explanation:
2Al + 2O2 → 2AlO + O2 Not Balanced Properly: 2Al + O2 = 2AlO
C2H4 + 3O2 → 2CO2 + 2H2O Looks Good
2CH4 + O2 → 2CO + 4H2 Not Correct: CO should be CO2
Ca + O2 → CaOH Not Balanced and No source for the H
Answer:
1,300,000,000,000
Explanation:
1.3 x 10^12
We want to convert this from scientific notation.
Tip: in scientific notation the exponent tells you how many place you move the decimal point over. If the exponent is negative you move the decimal point to the left. Ex. For, 4.1 x 10^-8, we would move the decimal point over 8 times to the left to get .00000041. When the exponent is positive we move over to the right. Ex. For, 7.6 x 10^7 we would move the decimal point over 7 times to the right to get 76,000,000
So to convert 1.3 x 10^12 we simply move over the decimal point over 12 times to the right.
1.3 x 10^12 ------> 1,300,000,000,000
Our answer is 1,300,000,000,000
Answer:
The products are: A) CO2, H2O
Explanation:
Those products that are seen on the right side of the reaction (that is, those substances that are generated from the reagents). In this case they are carbon dioxide and water.
The general equation of cellular respiration is:
C6H1206 + 602 -> 36 ATP + 6CO2 + 6H20
Answer:
Average atomic mass = 85.557 amu.
Explanation:
Given data:
Percent abundance of Rb-85 = 72.15%
Percent abundance of Rb-87 = 27.85%
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (72.15×85)+(27.85×87) /100
Average atomic mass = 6132.75 + 2422.95 / 100
Average atomic mass = 8555.7 / 100
Average atomic mass = 85.557 amu.
