T½=18.72days
therefore t¾=18.72+½ of 18.72
we have 18.72+9.36=28.08days
Answer:
f. Sn^4+
c. second
e. Al^3+
d. third
Explanation:
This question comes from a quantitative analysis showing the flowchart of a common scheme for identifying cations.
Now, from the separation scheme, Let's assume that Sn⁴⁺ & Al³⁺ were given; Then, Yes, the separation will work.
However, there will be occurrence of precipitation after the 1st step1.
So, the <u>Sn⁴⁺</u> cation will precipitate after the <u>second </u>step. Then the <u>Al³⁺</u> cation will precipitate after the <u>third</u> step.
Answer is: volume is 25.08 liters.
Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂(g).
m(Na) = 51.5 g; mass of sodium.
n(Na) = m(Na) ÷ M(Na).
n(Na) = 51.5 g ÷ 23 g/mol.
n(Na) = 2.24 mol; amount of sodium.
From chemical reaction: n(Na) : n(H₂) = 2 : 1.
n(H₂) = 2.24 mol ÷ 2.
m(H₂) = 1.12 mol; amount of hydrogen gas.
V(H₂) = n(H₂) · Vm.
V(H₂) = 1.12 mol · 22.4 L/mol.
V(H₂) = 25.08 L; volume of hydrogen gas.
Answer:
Chemical reaction involved is:
CaCO
3
→CaO+CO
2
Molar mass of CaCO
3
=40+12+16×3=100 g
According to the balanced reaction 100 g of CaCO
3
produces 1 mol of CO
2
i.e. 22.4 L at STP.
∴50 g CaCO
3
will produce
100
22.4
×50=11.2 L CO
2
at STP.
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