Answer:
dispersion forces
Explanation:
SO3 is a trigonal planar molecule. All the dipoles of the S-O bonds cancel out making the molecule to be a nonpolar molecule.
The primary intermolecular force in nonpolar molecules is the London dispersion forces. As expected, the London dispersion forces is the intermolecular force present in SO3.
Hence SO3 is a symmetrical molecule having only weak dispersion forces acting between its molecules.
Answer:
THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.
Explanation:
Equation:
Al3+ + 3e- -------> Al
3 F of electricity is required to produce 1 mole of Al
3 F of electricity = 27 g of Al
If 18 g of aluminium was used, the quantity of electricity to be used up will be:
27 g of AL = 3 * 96500 C
18 G of Al = x C
x C = ( 3 * 96500 * 18 / 27)
x C = 193 000 C
For 18 g of Al to be produced, 193000 C of electricity is required.
To calculate the current required to produce 193 000 C quantity of electricity, we use:
Q = I t
Quantity of electricity = Current * time
193 00 = I * 1.50 * 60 * 60 seconds
I = 193 000 / 1.50 * 60 *60
I = 193 000 / 5400
I = 35.74 A
The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A
Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
or,
where,
= initial pressure = 3.7 atm
= final pressure = ?
= initial volume = 5.8 L
= final volume = 2.5 L
Now put all the given values in the above equation, we get:
Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.
18.4 * 10^9 lbs = (18.4 * 10^9)/2000 tons
Cost = ((18.4 * 10^9)/2000 ) * 318
Cost = $ 2925600000
Your answer is right.
Important elements to consider:
- to use the balanced equation (which you did)
- divide the masses of each compound by the correspondant molar masses (which you did)
- compare the theoretical proportions with the current proportions
Theoretical: 2 mol of Na OH : 1 mol of CuSO4
Then 4 mol of NaOH need 2 mol of CUSO4.
Given that you have more than 2 mol of of CUSO4 you have plenty of it and the NaOH will consume first, being this the limiting reagent.
