Answer:
check the attached image. hope it helps
Explanation:
a. What does mol/L mean? mol/L means Molar Concentration
b. Describe in your own words how mol/L compares to grams/liter. (similarities and differences)
<u>Similarities:</u><u> </u>mol/L and grams/liter both are units of concentration.
<u>Differences</u>:
mol/L concentration says how many moles of solute are present in 1L of solution.
grams/liter concentration says how many grams of solute are present in 1L of solution.
3. Describe at least 2 ways in the simulation to change each of the parameters:
a. Volume of solution= evaporating or adding more water
b. amount of solute=reducing the water or evaporating more water
c. Concentration of solute in solution = evaporating or adding more solute
Answer:
0.382g
Explanation:
Step 1: Write the reduction half-reaction
Al³⁺(aq) + 3 e⁻ ⇒ Al(s)
Step 2: Calculate the mass of Al produced when a current of 100. A passes through the cell for 41.0 s
We will use the following relationships.
- 1 mole of electrons has a charge of 96486 C (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons pass through the cell.
- The molar mass of Al is 26.98 g/mol.
The mass of Al produced is:
Answer:
27 liters of hydrogen gas will be formed
Explanation:
Step 1: Data given
Number of moles C = 1.03 moles
Pressure H2 = 1.0 atm
Temperature = 319 K
Step 2: The balanced equation
C +H20 → CO + H2
Step 3: Calculate moles H2
For 1 mol C we need 1 mol H2O to produce 1 mol CO an 1 mol H2
For 1.03 moles C we'll have 1.03 moles H2
Step 4: Calculate volume H2
p*V = n*R*T
⇒with p = the pressure of the H2 gas = 1.0 atm
⇒with V = the volume of H2 gas = TO BE DETERMINED
⇒with n = the number of moles H2 gas = 1.03 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 319 K
V = (n*R*T)/p
V = (1.03 * 0.08206 *319) / 1
V = 27 L
27 liters of hydrogen gas will be formed
Answer:
C2H2O4
Explanation:
To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :
C = 26.86%
H = 2.239%
O = 100 - ( 26.86 + 2.239) = 70.901%
We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u
The division is as follows:
C = 26.86/12 = 2.2383
H = 2.239/1 = 2.239
O = 70.901/16 = 4.4313
We now divide each by the smallest number I.e 2.2383
C = 2.2383/2.2383 = 1
H = 2.239/2.2383 = 1
O = 4.4313/2.2383 = 1.98 = 2
Thus, the empirical formula is CHO2.
To get the molecular formula, we use the molar mass .
(CHO2)n = 90
We add the atomic masses multiplied by n.
(12 + 1 + 2(16))n = 90
45n = 90
n = 90/45 = 2.
Thus , the molecular formula is C2H2O4
