Answer:
Oceanic trenches
Explanation:
Oceanic Trenches are hemispheric-scale long but narrow topographic depressions of the sea floor
Answer: Explanation:
When supply and demand for a product increase simultaneously, we cannot predict the market clearing price, but know that the equilibrium quantity will increase.
NB: We cannot predict the market clearing price because no adequate information such as government policy is provided. such information would likely affect the market clearing price either positively or negatively.
P.S: This is a business question.
Answer:
Formula: H2O Formula Weight: 18.02 CAS No.: 7732-18-5 Density: 1.000 g/mL at 3.98 °C(lit.)
Explanation:
Answer:
a) Acidic buffer
b) No buffer
c) Acidic buffer
d) Basic buffer
e) Basic buffer
Explanation:
a) 75.0 mL of 0.10 M HF ; 55.0 mL of 0.15 M NaF
-Acidic buffer
Mixing of 75.0 mL of 0.10 HF and 55.0 mL of 0.15 mL NaF results in acidic buffer. HF/NaF is a buffer of weak acid and its conjugate base. F- is the conjugate base of acid,HF.
b.) 150.0 mL of 0.10 M HF ; 135.0 mL of 0.175 M HCl-No buffer
Mixing HF and HCl will not results in a buffer. Both are acids, and no conjugate base is present.
c.) 165.0 mL of 0.10 M HF ; 135.0 mL of 0.050 M KOH-Acidic buffer
HF reacts with KOH to form KF. F- is a conjujate base of HF. As volume and concentration of HF is more than KOH, therefore, HF will remain after reaction with KOH. HF/KF will be a buffer of weak acid and its conjugate base.
d.) 125.0 mL of 0.15 M CH3NH2 ; 120.0 mL of 0.25 M CH3NH3Cl -Basic buffer
CH3NH2/CH3NH3+ is a buffer of weak base and its conjugate acid.
e.) 105.0 mL of 0.15 M CH3NH2 ; 95.0 mL of 0.10 M HCl-Basic buffer
CH3NH2 is a weak base and HCl is a strong acid. CH3NH2 reacts with HCl to form its conjugate acid CH3NH3+. Volume and concentration of CH3NH2 is more as compared to HCl and hence, will remain in the soution after reactionf with HCl.
CH3NH3+/CH3NH2 is a buffer of weak base and its conjugate acid.
Heat gained by ice cubes would be equal to the - heat lost by warm water
The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol
Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J
Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C
In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:
q(ice/water) = - q(warm water)
moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)
50.5 g / 18.0 g/mol = 2.81 mol
2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)
16916 + 211.3T2 = -669.4 T2 + 53555
36639 = 880.7 T2
T2 = 41.6 C
