Answer:
5400 cans
Explanation:
First we convert the total weight, 1 ton, to grams:
Now we need to know the mass of aluminum:
Now we make the relation between the mass of aluminum in 1 ton of the earth's crust and the mass of aluminum per can:
It would be 1 or 2 because if the number is higher than 5 you need to round up , if its lower than 5 you need to round down.
moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
Answer:
Mass = 135.66 ×10⁻²¹ g
Explanation:
Given data:
Number of molecules of CuSO₄= 5.119×10²
Mass of CuSO₄= ?
Solution:
The given problem will solve by using Avogadro number.
1 mole contain 6.022×10²³ molecules
5.119×10² molecules ×1 mol / 6.022×10²³ molecules
0.85×10⁻²¹ mol
Mass in grams:
Mass = number of moles × molar mass
Mass = 0.85×10⁻²¹ mol × 159.6 g/mol
Mass = 135.66 ×10⁻²¹ g
