Answer:
2.956 moles chlorine gas will be produced
Explanation:
Step 1: data given
Number of moles carbon tetrachloride (CCl4) = 0.739 moles
carbon disulfide (s) = CS2(s)
chlorine (g) = Cl2(g)
carbon tetrachloride (l) = CCl4(l)
sulfur dichloride (s) = SCl2 (s)
Step 2: The balanced equation
CS2(s) + 4Cl2(g) → CCl4(l) +2SCl2
Step 3: Calculate moles chlorine gas
For 1 moles Cs2 we need 4 moles Cl2 to produce 1 mol CCl4 and 2 moles SCl2
For 0.739 moles CCl4 we need 4*0.739 = 2.956 moles Cl2
2.956 moles of chlorine gas will be produced
C5H12 (l) + 8O2 (g) ----> 5CO2 (g) + 6H2O (l)
Delta H = -3505.8 kJ/mol
C (s) + O2 (g) -----> CO2 (g)
Delta H = -393.5 kJ/mol
H2 (g) + (1/2)O2 (g) ------> H2O (l)
Delta H = -286 kJ/mol
Possible answers:
a. +35 kJ/mol
b. + 1,073 kJ/mol
c. -4,185 kJ/mol
d. -2,826 kJ/mol
e. -178 kJ/mol
Answer:
10−8 M.
Explanation:
In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
Answer:
A. K
Step-by-step explanation:
Remember the trends in the Periodic Table:
- Atomic radii <em>decrease</em> from left to right across a Period.
- Atomic radii <em>increase</em> from top to bottom in a Group.
- Ionic radii of metal cations are <em>smaller</em> than those of their atoms.
Thus, the largest atoms are in the lower left corner of the Periodic Table.
The diagram below shows that K is closest to the lower left, so it is the largest atom. It is also larger than any of the cations.