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Alika [10]
3 years ago
6

A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbon

ate that has twice the mass of the first sample
Chemistry
1 answer:
Anna35 [415]3 years ago
3 0

Answer:

Percentage lithium by mass in Lithium carbonate sample = 19.0%

Explanation:

Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g

Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g

Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%

Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g

Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%

Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g

mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g

Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%

Percentage lithium by mass in Lithium carbonate sample = 19.0%

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cupoosta [38]

Answer:

m_{CaCO_3}=0.179gCaCO_3

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3

Regards.

3 0
4 years ago
What is the empirical formula for a compound that is 94.1% oxygen and 5.90 % hydrogen?
RSB [31]

Answer:

The empirical formula would be N₂Os * Page 2 Calculate the empirical formulaof a compound that is 94.1% oxygen, 5.9% hydrogen.

6 0
3 years ago
The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate
Nimfa-mama [501]

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8 cm

for a BCC structure, the atomic radius is equal to

r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3}  }{4}=2.3103x10^{-8}cm

7 0
3 years ago
Using your data above, draw conclusions about the d-splitting for each ligand (H2O, en, phen). Order the complexes from least to
Delvig [45]

Answer:

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7 0
3 years ago
What statements are always true about limiting reactants?
alisha [4.7K]

Answer:

5.The limiting reactant is completely used up in the reaction

Explanation:

The limiting reactant is completely used up in the reaction is the correct answer because a limiting reactant is a reactant in chemical reaction that is completely consumed or used up in the chemical reaction. Limiting reactant when it is completely used up limits the amount of products that will be formed. The reaction will be halted or will stop when the limiting reactant is totally used up.

7 0
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