Answer : The molecular weight of a gas is, 128.9 g/mole
Explanation : Given,
Density of a gas = 5.75 g/L
First we have to calculate the moles of gas.
At STP,
As, 22.4 liter volume of gas present in 1 mole of gas
So, 1 liter volume of gas present in
mole of gas
Now we have to calculate the molecular weight of a gas.
Formula used :

Now put all the given values in this formula, we get the molecular weight of a gas.


Therefore, the molecular weight of a gas is, 128.9 g/mole
Ribosomes<span> are connected to the ER membrane, which makes it “rough.” The Rough ER is connected to the nuclear envelope surrounding the nucleus as well.</span>
Answer:
A) ΔG° = -3,80x10⁵ kJ
B) E° = 2,85V
Explanation:
A) It is possible to answer this problem using the standard ΔG's of formation. For the reaction:
Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)
The ΔG° of reaction is:
ΔG° = ΔGFe(s) + ΔGMg²⁺(aq) - (ΔGFe²⁺(aq) + ΔGMg(s) <em>(1)</em>
Where:
ΔGFe(s): 0kJ
ΔGMg²⁺(aq): -458,8 kJ
ΔGFe²⁺(aq): -78,9 kJ
ΔGMg(s): 0kJ
Replacing in (1):
ΔG° = 0kJ -458,8kJ - (-78,9kJ + okJ)
<em>ΔG° = -3,80x10² kJ ≡ -3,80x10⁵ kJ</em>
B) For the reaction:
X(s) + 2Y⁺(aq) → X²⁺(aq) + 2Y(s)
ΔG° = ΔH° - (T×ΔS°)
ΔG° = -629000J - (298,15K×-263J/K)
ΔG° = -550587J
As ΔG° = - n×F×E⁰
Where n are electrons involved in the reaction (<em>2mol</em>), F is faraday constant (<em>96485 J/Vmol</em>) And E° is the standard cell potential
Replacing:
-550587J = - 2mol×96485J/Vmol×E⁰
<em>E° = 2,85V</em>
I hope it helps!
Answer:
Carbon and Hydrogen
Explanation:
So the C stands for Carbon and H is for Hydrogen. Methane (CH4) is made up of one atom of Carbon and 4 atoms of Hydrogen.
Hope this helps
<h3>
Answer:</h3>
54 g
<h3>
Explanation:</h3>
- The reaction between carbon and oxygen gas is given by the equation;
C(s) + O₂(g) → CO₂(g)
We are given;
18 g of Carbon
72 g of Carbon dioxide
- We need to calculate the amount of oxygen needed for the reaction.
- From the law of conservation of mass in chemical equation, the mass of the reactants is equal to the mass of the product.
Therefore;
Mass of Oxygen gas + mass of the carbon = Mass of carbon dioxide
Therefore;
Mass of Oxygen gas = mass of carbon dioxide - mass of carbon
= 72 g - 18 g
= 54 g
Therefore, the mass of Oxygen needed would be 54 g