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zmey [24]
4 years ago
9

2. What is the volume of 0.8 moles of Chlorine gas at STP?

Chemistry
1 answer:
ivann1987 [24]4 years ago
5 0

Answer:

17.92 litres.

Explanation:

1 mole of a gas occupies 22.4 litres at STP.

So 0.8 moles occupies 0.8 * 22.4

= 17.92 litres.

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An archaeologist discovers an ancient wooden bowl. which isotope would most likely be used to determine its age?
Vesnalui [34]

Answer:

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Explanation:

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7 0
3 years ago
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C12H22O11 + 12O2 ---> 12CO2 + 11H2O
boyakko [2]

Answer:

Oxygen is the limiting reactant.

Explanation:

Based on the reaction:

C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

<em>1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.</em>

<em />

10.0g of sucrose (Molar mass: 342.3g /mol) are:

10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁

And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:

10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂

For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):

0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = <em>0.3504 moles of O₂</em>

As you have just 0.3125 moles of O₂, <em>oxygen is the limiting reactant.</em>

8 0
4 years ago
Does anyone know the answer to this​
Damm [24]
It’s Tetracarbon dioxide :)
6 0
3 years ago
How low the freezing point can be depressed depends on the
Rainbow [258]
Answer is c I believe
5 0
4 years ago
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Copper crystallizes in a face-centered cubic unit cell. the density of copper is 8.94 g/cm3. calculate the length of the edge of
Ilia_Sergeevich [38]
<span>361.4 pm is the length of the edge of the unit cell. First, let's calculate the average volume each atom is taking. Start with calculating how many moles of copper we have in a cubic centimeter by looking up the atomic weight. Atomic weight copper = 63.546 Now divide the mass by the atomic weight, getting 8.94 g / 63.546 g/mol = 0.140685488 mol And multiply by Avogadro's number to get the number of atoms: 0.140685488 * 6.022140857x10^23 = 8.472278233x10^22 Now examine the face-centered cubic unit cell to see how many atoms worth of space it consumes. There is 1 atom at each of the 8 corners and each of those atoms is shared between 8 unit cells for for a space consumption of 8/8 = 1 atom. And there are 6 faces, each with an atom in the center, each of which is shared between 2 unit cells for a space consumption of 6/2 = 3 atoms. So each unit cell consumes as much space as 4 atoms. Let's divide the number of atoms in that cubic centimeter by 4 to determine the number of unit cells in that volume. 8.472278233x10^22 / 4 = 2.118069558x10^22 Now calculate the volume each unit cell occupies. 1 cm^3 / 2.118069558x10^22 = 4.721280262x10^-23 cm^3 Let's get the cube root to get the length of an edge. (4.721280262x10^-23 cm^3)^(1/3) = 3.61426x10^-08 cm Now let's convert from cm to pm. 3.61426x10^-08 cm / 100 cm/m * 1x10^12 pm/m = 361.4 pm Doing an independent search for the Crystallographic Features of Copper, I see that the Lattice Parameter for copper at at 293 K is 3.6147 x 10^-10 m which is in very close agreement with the calculated amount above. And since metals expand and contract with heat and cold, I assume the slight difference in values is due to the density figure given being determined at a temperature lower than 293 K.</span>
4 0
3 years ago
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