Answer : The new pressure of the gas will be, 468.66 atm
Explanation :
Boyle's Law : This law states that pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
or,

where,
= initial pressure of the gas = 74 atm
= final pressure of the gas = ?
= initial volume of the gas = 190 ml
= final volume of the gas = 30 ml
Now we put all the given values in the above formula, we get the final or new pressure of the gas.


Therefore, the new pressure of the gas will be, 468.66 atm
The Lewis structure/diagram for CH2O (aka Formaldehyde) can be written in either of the following ways shown in the picture.
The dots represent electrons in the valence shell of the atom (the outermost shell). The green dots are electrons that belong to the Oxygen atom, the blue belong to the Carbon atom, and the pink belong to the Hydrogen atoms.
Covalent bonds are bonds between atoms where atoms share electrons with each other. Atoms bond because they obey the octet rule ( the rule states that most atoms of main-group elements tend to want 8 electrons in their valence shells).
Oxygen has 6 valence electrons, Carbon has 4, and Hydrogen has 1. H does not follow the octet rule, but C and O do, so the atoms are arranged in this way so that the O and C atoms have a full octet of electrons in their valence.
<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
Neutral charges and are found in the center? I'm not sure for the last answer
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⁻