Answer:
Empirical formula is Cr₂O₃.
Explanation:
Given data:
Percentage of Cr = 68.4%
Percentage of O = 31.6%
Empirical formula = ?
Solution:
Number of gram atoms of Cr = 68.4 / 52 = 1.3
2
Number of gram atoms of O = 31.6 / 16 = 1.98
Atomic ratio:
Cr : O
1.32/1.32 : 1.98/1.32
1 : 1.5
Cr : O = 1 : 1.5
Cr : O = 2(1 : 1.5)
Empirical formula is Cr₂O₃.
Answer:
At one atmosphere and twenty-five degrees Celsius, could you turn it into a liquid by cooling it down? Um, and the key here is that the triple point eyes that minus fifty six point six degrees Celsius and it's at five point eleven ATMs. So at one atmospheric pressure, there's no way that you're ever going to reach the liquid days. So the first part of this question is the answer The answer to the first part of a question is no. How could you instead make the liquid at twenty-five degrees Celsius? Well, the critical point is at thirty-one point one degrees Celsius. So you know, if you're twenty-five, if you increase the pressure instead, you will briefly by it, be able to form a liquid. And if you continue Teo, you know, increase the pressure eventually form a salad, so increasing the pressure is the second part. If you increase the pressure of co two thirty-seven degrees Celsius, will you ever liquefy? No. Because then, if you're above thirty-one point one degrees Celsius in temperature. You'LL never be able to actually form the liquid. Instead, you'LL only is able Teo obtain supercritical co too, which is really cool thing. You know, they used supercritical sio tu tio decaffeinated coffee without, you know, adding a solvent that you'LL be able to taste, which is really cool. But no, you can't liquefy so two above thirty-one degrees Celsius or below five-point eleven atmospheric pressures anyway, that's how I answer this question. Hope this helped :)
Answer:
positive H and negative S
Explanation:
For a reaction to be spontaneous, the absolute best combination is a negative Delta H and a positive Delta S. When they are both positive, the reaction is only spontaneous at higher temperatures. When they are both negative, the reaction is only spontaneous at lower temperatures. and again if a catalyst is added to the reaction, the activation energy is lowered because a lower-energy transition state is formed. The catalyst does not affect the energy of the reactants or products (and thus does not affect ΔG).
So from these discussions
Ea does not affect G value at all (whether +Ea or -Ea).
And for product to be formed the reaction should be spontaneous, where H is negative and S positive else the reaction will yield low product.
When oxygen and hydrogen combine they form a polar covalent bond
1. The molar mass of the unknown gas obtained is 0.096 g/mol
2. The pressure of the oxygen gas in the tank is 1.524 atm
<h3>Graham's law of diffusion </h3>
This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e
R ∝ 1/ √M
R₁/R₂ = √(M₂/M₁)
<h3>1. How to determine the molar mass of the gas </h3>
- Rate of unknown gas (R₁) = 11.1 mins
- Rate of H₂ (R₂) = 2.42 mins
- Molar mass of H₂ (M₂) = 2.02 g/mol
- Molar mass of unknown gas (M₁) =?
R₁/R₂ = √(M₂/M₁)
11.1 / 2.42 = √(2.02 / M₁)
Square both side
(11.1 / 2.42)² = 2.02 / M₁
Cross multiply
(11.1 / 2.42)² × M₁ = 2.02
Divide both side by (11.1 / 2.42)²
M₁ = 2.02 / (11.1 / 2.42)²
M₁ = 0.096 g/mol
<h3>2. How to determine the pressure of O₂</h3>
From the question given above, the following data were obtained:
- Volume (V) = 438 L
- Mass of O₂ = 0.885 kg = 885 g
- Molar mass of O₂ = 32 g/mol
- Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
- Temperature (T) = 21 °C = 21 + 273 = 294 K
- Gas constant (R) = 0.0821 atm.L/Kmol
The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side by V
P = nRT / V
P = (27.65625 × 0.0821 × 294) / 438
P = 1.524 atm
Learn more about Graham's law of diffusion:
brainly.com/question/14004529
Learn more about ideal gas equation:
brainly.com/question/4147359
