Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
Answer:
0.479 M or mol/L
Explanation:
So Molarity is moles/litres of solution...often written as M=mol/L
So here we are given grams of BaCl2 which we have to convert to moles. To convert to moles of BaCl2 we have to divide 63.2 g BaCl2 by molar mass of BaCl2 which is 208.23 g/mol so you get 63.2/208.23 = 0.3035 moles of BaCl2
Second step is converting the 634mL to litres by simply dividing by 1000 because we know 1 litre has 1000ml so 634/1000 = 0.634L
Now we just plug these guys in our molarity formula M=mol/L
M= 0.3035/0.634 = 0.479 M or mol/L
Answer:
19.8 kg of C₂H₂ is needed
Explanation:
We solve this by a rule of three:
If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene
95.5×10⁴ kJ of heat may be released by the combustion of
(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂
Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g
If we convert the mass from g to kg → 19848 g . 1kg / 1000g = 19.8 kg
Answer:
C.
Explanation:
If the students want to know at what percent of CO2 in the air the plant will grow at the fastest, then the percent of CO2 should be a different value for each plant in the table.
There are 2 tables that have different values for the CO2 - the tables in answer choices C and D.
Since the students only want to know how the amount of CO2 affects the plant, every other variable should remain constant.
The only answer choice that has a changing value for the percent of CO2 and a constant value for every other variable is C.
Explanation:
Volume is the quantity of three-dimensional
