Answer:
b) +2 and +3.
Explanation:
Hello,
In this case, given the molecular formulas:
And:
We can relate the subscripts with the oxidation states by knowing that they are crossed when the compound is formed, for that reason, we notice that oxygen oxidation state should be -2 for both cases and the oxidation state of X in the first formula must be +2 since both X and O has one as their subscript as they were simplified:
Moreover, for the second case the oxidation state of X should be +3 in order to obtain 3 as the subscript of oxygen:
Thus, answer is b)+2 and +3
Best regards.
Answer:
20L
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 10L
Initial Temperature (T1) = stp = 273K
Final temperature (T2) = 546K
Final volume (V2) =..?
The new volume of the gas can be obtained by using the Charles' law equation as shown below:
V1/T1 = V2/T2
10/273 = V2/546
Cross multiply to express in linear form
273 x V2 = 10 x 546
Divide both side by 273
V2 = (10 x 546) / 273
V2 = 20L
Therefore, the new volume of the gas is 20L.
<h3>
Answer:</h3>
B. 0.33 mol
<h3>
Explanation:</h3>
We are given;
Gauge pressure, P = 61 kPa (but 1 atm = 101.325 kPa)
= 0.602 atm
Volume, V = 5.2 liters
Temperature, T = 32°C, but K = °C + 273.15
thus, T = 305.15 K
We are required to determine the number of moles of air.
We are going to use the concept of ideal gas equation.
- According to the ideal gas equation, PV = nRT, where P is the pressure, V is the volume, R is the ideal gas constant, (0.082057 L.atm mol.K, n is the number of moles and T is the absolute temperature.
- Therefore, to find the number of moles we replace the variables in the equation.
- Note that the total ball pressure will be given by the sum of atmospheric pressure and the gauge
- Therefore;
- Total pressure = Atmospheric pressure + Gauge pressure
We know atmospheric pressure is 101.325 kPa or 1 atm
Total ball pressure = 1 atm + 0.602 atm
= 1.602 atm
That is;
PV = nRT
n = PV ÷ RT
therefore;
n = (1.602 atm× 5.2 L) ÷ (0.082057 × 305.15 K)
= 0.3326 moles
= 0.33 moles
Therefore, there are 0.33 moles of air in the ball.
<span>We can use
the heat equation,
Q = mcΔT </span>
<span>Where Q is
the amount of energy transferred (J), m is the mass of the substance
(kg), c is
the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is
the temperature difference (°C).</span>
Let's assume that the finale temperature is T.
Q = 1200 J
<span>
m = 36 g
c = 4.186 J/g °C</span>
ΔT = (T -
22)
By applying
the formula,
1200 J = 36 g
x 4.186 J/g °C x (T - 22)
(T - 22) = 1200 J / (36 g x 4.186 J/g °C)
(T - 22) = 7.96 °C
T = (7.96 + 22) °C = 29.96 °C
T = 30 °C
Hence,
the final temperature is 30 °C.
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
