Answer:
Final concentrations:
Cu²⁺ = 0
Al³⁺ = 3.13 mmol/L = 84.51 mg/L
Cu = 4.7 mmol/L = 300 mg/L
Al = 0.57 mmol/L = 15.49 mg/L
Explanation:
2Al (s) + 3Cu²⁺ (aq) → 2Al³⁺ (aq) + 3Cu (s)
Al: 27 g/mol ∴ 100 mg = 3.7 mmol
Cu: 63.5 g/mol ∴ 300 mg = 4.7 mmol
3 mol Cu²⁺ _______ 2 mol Al
4.7 mmol Cu²⁺ _____ x
x = 3.13 mmol Al
4.7 mmol of Cu²⁺ will be consumed.
3.13 mmol of Al will be consumed.
4.7 mmol of Cu will be produced.
3.13 mmol of Al³⁺ will be produced.
0.57 mmol of Al will remain.
Answer:
d = 43.5 g/cm³
Explanation:
Given data:
Mass of magnesium cube = 217.501 g
Volume of magnesium cube = 5.00 cm³
Density of magnesium cube = ?
Solution:
Formula:
d = m/v
d = density
m = mass
v = volume
by putting values,
d = 217.501 g/ 5.00 cm³
d = 43.5 g/cm³
When the reaction equation is:
CaSO3(s) → CaO(s) + SO2(g)
we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.
to get the moles of SO2 we are going to use the ideal gas equation:
PV = nRT
when P is the pressure = 1.1 atm
and V is the volume = 14.5 L
n is the moles' number (which we need to calculate)
R ideal gas constant = 0.0821
and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K
so, by substitution:
1.1 * 14.5 L = n * 0.0821 * 285.5
∴ n = 1.1 * 14.5 / (0.0821*285.5)
= 0.68 moles SO2
∴ moles CaSO3 = 0.68 moles
so we can easily get the mass of CaSO3:
when mass = moles * molar mass
and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol
∴ mass = 0.68 moles* 120 g/mol = 81.6 g
Answer:
i mean kinda they should throw people in the air tho lol
Explanation:
<u>Answer:</u> The volume of stock solution needed is 90 mL
<u>Explanation:</u>
To calculate the molarity of the diluted solution, we use the equation:
where,
are the molarity and volume of the stock sulfuric acid solution
are the molarity and volume of diluted sulfuric acid solution
We are given:
Putting values in above equation, we get:
Hence, the volume of stock solution needed is 90 mL
