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
The chemical compound's empirical formula is NS.
The chemical compound's molecular formula is N4S4.
<h3>What does a chemical empirical formula look like?</h3>
- The empirical formula of a compound that gives the proportion (ratios) of the elements in the complex but not the precise number or arrangement of atoms is known as an empirical formula.
- This would be the compound's element to whole number ratio with the lowest value.
<h3>What sort of empirical formula would that be?</h3>
- The chemical structure of glucose is C6H12O6. Every mole of carbon and oxygen is accompanied by two moles of hydrogen.
- Glucose has the empirical formula CH2O.
- Ribose has the chemical formula C5H10O5, which can be simplified to the empirical formula CH2O.
learn more about empirical formula here
brainly.com/question/1603500
#SPJ4
the question you are looking for is
A compound containing only sulfur and nitrogen is 69.6% S by mass; the molar mass is 184 g/mol. What are the empirical and molecular formulas of the compound?
Answer:
The answer to your question is: 6.8 g of water
Explanation:
Data
2.6 moles of HCl
1.4 moles of Ca(OH)2
2HCl + Ca(OH)2 → 2H2O + CaCl2
MW 2(36.5) 74 36 g 111 g
73g
1 mol of HCl ---------------- 36.5 g
2.6 mol -------------- x
x = (2.6 x 36.5) / 1 = 94.9 g
1 mol of Ca(OH)2 -------------- 74 g
1.4 mol --------------- x
x = (1.4 x 74) / 1 = 103.6 g
Grams of water
73 g of HCl ------------------ 36g of H2O
94.9 g ------------------- x
x = (94.9 x 36) / 73 = 46.8 g of water
