Dipole-dipole interactions, and London dispersion interactions
Answer: This can be quickly solved with "traintracks"
Explanation:
You start w/ grams of water and want to find moles of oxygen gas produced.
So you want to Convert:
Grams of water -> moles of water -> moles of oxygen gas.
The two things you need to know to set up the tracks are:
1)Molar mass of water- H2O
Hydrogen - 1.008(x2)
Oxygen - 16.00
Water - 18.016
The sample has a new pressure of 274kPa. If at 105 kPa and 275K, a 220 mL sample of helium gas is contained in a cylinder with a moving piston. The sample is pushed till it has a 95.0 mL volume and 310K .
The macroscopic characteristics of ideal gases are related by the ideal gas law (PV = nRT). A gas is considered to be perfect if its particles (a) do not interact with one another and (b) occupy no space (have no volume). Where P= pressure V= volume and T = temperature.
From ideal gas equation
P₁V₁/T₁ =P₂V₂/T₂
105×220÷275 = P₂ ×95÷310
P₂= (105×220×310)÷(275×95)
P2= 7161000/26125
P2 = 274.105 kPa
Hence, the new pressure of helium gas is 274kPa
To know more about Ideas gas equation
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Answer:
the spotlight effect.
Explanation:
The spotlight effect is a tendency to think that people get noticed more often than they really do. It is an overestimation of the situation regarding the concern of getting observed. It concerns the self-confidence of an individual. For example, an individual feels that everybody in a party would notice him for a bad pair of shoes while in reality, it does not concern them.
As per the question, the overestimation of people's reaction is known as the spotlight effect.
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻