1. P = F/A; weight is a force (the force of gravity on an object), so divide the weight by the area given. P = 768 pounds/75.0 in² = 10.2 pounds/in².
2. Using the same equation from question 1, rearrange it to solve for A: A = F/P. We're given the force (the weight) and the pressure, so A = 125 pounds/3.25 pounds/in² = 38.5 in².
3. Again, using the same equation from question 1, rearrange it this time to solve for F: F = PA = (4.33 pounds/in²)(35.6 in²) = 154 pounds.
4. We can set up a proportion given that 14.7 PSI = 101 KPa. This ratio should hold for 23.6 PSI. In other words, 14.7/101 = 23.6/x; to solve for x, which would be your answer, we compute 23.6 PSI × 101 kPa ÷ 14.7 PSI = 162 kPa.
5. We are told that 1.00 atm = 760. mmHg, and we want to know how many atm are equal to 854 mmHg. As we did with question 4, we set up a proportion: 1/760. = x/854, and solve for x. 854 mmHg × 1.00 atm ÷ 760. mmHg = 1.12 atm.
6. The total pressure of the three gases in this container is just the sum of the partial pressures of each individual gas. Since our answer must be given in PSI, we should convert all our partial pressures that are not given in PSI into PSI for the sake of convenience. Fortunately, we only need to do that for one of the gases: oxygen, whose partial pressure is given as 324 mmHg. Given that 14.7 PSI = 760. mmHg, we can set up a proportion to find the partial pressure of oxygen gas in PSI: 14.7/760. = x/324; solving for x gives us 6.27 PSI oxygen. Now, we add up the partial pressures of all the gases: 11.2 PSI nitrogen + 6.27 PSI oxygen + 4.27 PSI carbon dioxide = 21.7 PSI, which is our total pressure.
Answer:
The ratio of pressure in bottle B to that of bottle A is 1 : 4
Explanation:
We'll be by calculating the pressure in both bottles. This is illustrated below below:
For A:
Temperature (T) = T
Volume (V) = V
Number of mole (n) = n
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =...?
PV = nRT
PV = n x 0.0821 x T
Divide both side by V
P = nT0.0821/V
Therefore, the pressure, in bottle A is
PA = nT0.0821/V
For B:
Temperature (T) = the same as that of A = T
Volume (V) = twice that of A = 2V
Number of mole (n) = half that of A = ½n
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =...?
PV = nRT
P x 2V = ½n x 0.0821 x T
Divide both side by 2V
P = ½n x 0.0821 x T/2V
P = nT0.0821/4V
Therefore, the pressure in bottle B is:
PB = nT0.0821/4V
Now, we can obtain the ratio of pressure in bottle B to that of bottle A as follow:
Pressure in bottle A (PA) = nT0.0821/V
Pressure in bottle B (PB) = nT0.0821/4V
PB/PA = nT0.0821/4V ÷ nT0.0821/V
PB/PA = nT0.0821/4V x V/nT0.0821
PB/PA = 1/4
Therefore, the ratio of pressure in bottle B to that of bottle A is 1 : 4.
Answer:
1.13 × 10⁶ g
Explanation:
Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.
Al³⁺ + 3 e⁻ → Al
We can establish the following relations:
- 1 Ampere = 1 Coulomb / second
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons circulate
- The molar mass of Al is 26.98 g/mol.
The mass of aluminum produced under these conditions is:
Missing question:
a. C6H12O6
<span>b. CH3OH </span>
<span>c. C12H22O11 </span>
<span>d. NaOH.
</span>Answer is: d. NaOH.
b(solution) = 0,1 m.
ΔT = Kf ·
b(solution) · i.
Kf - the freezing point depression constant.
i - Van 't Hoff factor. Because dissociate on one cation and one anions, sodium hydroxide has i = 2, other molecules has covalent bonds (i = 1, do not dissociate on ions).
Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.
