This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.
Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O
The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.
MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄
Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.
4 protons are there in the atom
The pressure of 1.27 L of a gas at 288°C, if the gas had a volume of 875 ml at 145 kPa and 176°C is 1.195 atm.
<h3>What is ideal gas equation?</h3>
Ideal gas equation of any gas will be represented as:
PV = nRT, where
P = pressure
V = volume
n = moles
R = universal gas constant
T = temperature
First we calculate the moles of gas, when the volume of gas 875 ml at
145 kPa and 176°C as:
n = (1.431atm)(0.875L) / (0.082L.atm/K.mol)(449.15K)
n = 1.252 / 36.83 = 0.033 moles
Now we measure the pressure of 0.033 moles of gas of 1.27 L of a gas at 288°C as:
P = (0.033mol)(0.082L.atm/K.mol)(561K) / (1.27L) = 1.195 atm
Hence required pressure of gas is 1.195 atm.
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Answer:
R=0.0438 Ω
Explicación:
1) Hallar el área o sección del conductor de cobre, usando esta fórmula:
A=π.r² (Pi x radio al cuadrado)
Debido a que conocemos el diámetro (1.5mm) su radio es la mitad de esto es decir 0.75mm, y lo sustituimos en la fórmula:
A=π.(0.75mm)²
A=π(0.5625mm²)
A=1.7671mm²
2) La resistividad del cobre es: rho = 0,0172 y la incluimos en la fórmula siguiente:
R=p
R=0,0172Ω x
Simplificamos:
R=
El resultado es:
R=0.0438 Ω
Explanation:
