Answer: The volume of the gas at a pressure of 800 mm Hg is 75 mL.
Explanation:
Given: = 100.0 mL,
= 600 mm Hg
= ?,
= 800 mm Hg
According to Boyle's law, at constant temperature the pressure of given mass of an ideal gas is inversely proportional to volume.
Hence, formula used to calculate new volume of the gas is as follows.
Substitute the values into above formula.
Therefore, we can conclude that the volume of the gas at a pressure of 800 mm Hg is 75 mL.
Answer:
The structures are shown below
Explanation:
To draw the Lewis structure, we must place the central atom with its valence electrons surrounded by the other atoms, with their valence electrons, then, using lines, we put the bonds (simple line for 1 pair sharing, a double line for 2 pairs sharing, and triple line for 3 pairs sharing).
For N₂H₄, the two nitrogens are bonded together and with two hydrogens, Nitrogen has 5 valence electrons, and hydrogen 1 valence electron, so, 3 bonds will be done by nitrogen, so there will one lone pair in the structure for each nitrogen, and none for hydrogen. The structure is shown below.
For the N₂H₂, the nitrogens are bonded together, sharing two pairs of electrons, and are each one is bonded to a hydrogen, sharing one pair of electrons. Both nitrogen, then, has one lone pair of electron. The structure is shown below.
For the N₂, the two nitrogens are sharing three pairs of electrons, and both have one lone pair of electron. The structure is shown below.
The answer is:
The new pressure will be equal to 6 atm.
Since we know that the temperature is constant, we can use the Boyle's Law to solve the problem.
The Boyle's Law establishes that when the temperature is kept constant, the pressure and the volume will be proportional.
So, we have the equation:
We are given,
Substituting and isolating from the Boyle's Law, we have:
Hence, the new pressure will be equal to 6 atm.
Have a nice day!
Answer:
Given nuclide is uranium-234.
If it undergoes beta decay then, what will be the daughter particle?
Explanation:
The beta particle decay increases the atomic number of the nuclide by one.
There is no change in the mass number of the nuclide.
The decay of beta particle from Uranium-234 is shown below:
It forms Np-93.
