We have to calculate the percentage (by mass) of Ba in the mixture of BaBr₂ and inert material.
Given, bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate as per following reaction:
BaBr₂(aq) + 2AgNO₃(aq)→2AgBr(s)+Ba(NO₃)₂(aq). From this balanced chemical reaction, it is clear that one mole BaBr₂ reacts with two moles of AgNO₃ and gives two moles of AgBr. We have to calculate 0.6226 g AgBr contains how many moles of AgBr. Using that information we can get how many moles of BaBr₂ reacted to give 0.6226g AgBr and one mole BaBr₂ contains one mole of Ba. By multiplying number of moles of Ba with atomic mass of Ba we can get amount of Ba present in the mixture. Accordingly we can calculate mass percentage of Ba in the mixture.
Atomic mass of Ba= 137.327 g.mol, Molecular mass of BaBr₂=297.1 g/mol and molecular mass of AgBr=187.7 g/mol. Mass of AgBr is 0.6226 g which contains 0.6226/187.7 mole= 3.31X10⁻³ moles of AgBr. So, moles of BaBr₂ reacts= (3.31X10⁻³)/2 moles= 1.65X10⁻³ moles. One mole BaBr₂ contains one mole of Ba. So, 1.65X10⁻³ moles of BaBr₂ contains 1.65X10⁻³ moles of Ba atoms whose mass is=1.65X10⁻³X137.327g=0.2265 g. Out of 0.7207 g sample amount of Ba present is 0.2265 g. So, mass percentage is =31.42 %
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Answer:Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance; equivalently, it is the ratio of the mass of a substance to the mass of a reference substance for the same given volume.
Explanation:
Answer:
3.39g of MgBr
Explanation:
Reaction between magnesium and bromine
Mg + Br₂ → MgBr
Mass of MgBr = ?
1g of Mg + 5.0g of Br₂
6g of Mg and Br₂
Molar mass of Mg = 24g/mol
Molar mass of Br = 80g/mol
Number of moles = mass / molar mass
Mass = number of moles * molarmass
Mass of Mg = 1 * 24 = 24g
Mass of Br₂ = 1 * (2*80) = 160g (diatomic molecule)
Molarmass of MgBr = (24 + 80) = 104g/mol
Mass of MgBr = 1 * 104 = 104g/mol
24g of Mg + 160g of Br₂ = 104g of MgBr
184g of (mg and Br) = 104g of mgBr
6g of (mg and Br) = y g of MgBr
y = (6 * 104) / 184
y = 3.39g of MgBr
Your answer is D. 8.5 x 10^12. Reason is because if you use a calculator (ex. TI-84 Plus C Silver Edition Calculator), it will say 8.5E12, which is represented 8.5 x 10^12.
Hope this helped!
Nate
Answer:
At the higher altitude, the new volume is 750.2L
Explanation:
If we decompose the Ideal Gases Law for the two situations (initial and final), we can determine this relaion:
P₁ . V₁ / T₁ = P₂ . V₂ / T₂
Number of moles does not change, and R stays the same.
Let's make some conversions, before:
752 mmHg . 1 atm / 760mmHg = 0.989 atm
24.3°C + 273 = 297.3K
0.0708 bar . 0.986 atm / 1bar = 0.070 atm
-5.41°C + 273 = 267.59 K
We replace: (0.989 atm . 59L) / 297.3K = (0.070 atm . V₂) / 267.59K
[(0.989 atm . 59L) / 297.3K] . 267.59K = 0.070 atm . V₂
52.5 atm.L = 0.070 atm . V₂
V₂ = 52.5 atm.L / 0.070atm = 750.2L