Answer:
D
Explanation:
Color, texture, and density are all physical properties but reactivity is a chemical property so the answer is D.
Answer:
b) 18.5 g
Explanation:
The first step is the <u>calculation of number of moles</u> of 
with the molarity equation, so:
With the <u>volume in "L" units</u> (342.6 mL= 0.342 L ) we can <u>find the moles</u>, so:
Now, we have to calculate the <u>molar mass</u> to convert the moles to grams. For this we have to know the <u>atomic mass</u> of each atom:
Li ( 6.94 g/mol)
N (14 g/mol
O (16 g/mol)
With these values and the number of atoms in the molecule we can do the math:
(6.94 x 1) + ( 14 x 1) + (16 x 3 ) = 68.94 g/mol
The final step is the <u>calculation of the grams</u>, so:
<u>We will need 18.5 g of LiNO3 to do a 0.783 M solution with a volume 342.6 mL.</u>
Answer:
x= 138.24 g
Explanation:
We use the avogradro's number
6.023 x 10^23 molecules -> 1 mol C2H8
26.02 x 10^23 molecules -> x
x= (26.02 x 10^23 molecules * 1 mol C2H8 )/6.023 x 10^23 molecules
x= 4.32 mol C2H8
1 mol C2H8 -> 32 g
4.32 mol C2H8 -> x
x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8
x= 138.24 g
Answer:
Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.
Explanation:
- It is a stichiometry problem.
- We should write the balance equation of the mentioned chemical reaction:
<em>2Al + 3CuCl₂ → 3Cu + 2AlCl₃.</em>
- It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.
- Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:
<em>n = mass / molar mass</em>
- The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.
- The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.
- <em>From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.</em>
∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with <em>(2:3)</em> ratio and CuCl₂ is the limiting reactant while Al foil is in excess.
- From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
- So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
- Finally, we can calculate the mass of copper produced using:
mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.
- <u><em>So, the answer is:</em></u>
<em>Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.</em>
According to Graham's law, the rate of effusion of a gas is inversely proportional to square root of its molecular weight.
This can be represented as follows.
Rate of effusion ∝ 1/√M
Therefore to find which gas has highest rate of effusion, we will find out the molar masses of the given compounds. The gas that is lighter in weight would have highest rate of effusion.
1) Molar mass of oxygen (O₂) is 32 g and that of H₂ is 2.01 g . Therefore H₂ would have highest rate of effusion
2) Molar mass of methane is 16.05 [12.01 + 4 (1.01)] g and that of CCl₄ [ 12 + 4(35.45) ] is 154 g. Therefore methane will have highest rate of effusion
3) Molar mass of N₂ is 28 g and molar mass of NH₃ is [ 14 + 3(3.01) ] = 17.03 g.
Therefore NH₃ will have highest rate of effusion.
4) Molar mass of F₂ is 38 g and that of Cl₂ is 71 g. Therefore F₂ will have highest rate of effusion
