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Alex
3 years ago
8

Which is not a property covalent bonds ?

Chemistry
1 answer:
viktelen [127]3 years ago
4 0

your answer would be a. usually liquid or gas at room temperature

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What is the limiting reactant when 19.9 g CuO react with 2.02 g H2?
Harlamova29_29 [7]

Answer:

Explanation:

use the equation

moles = mass/mr

=19.9/79.5

=0.250moles of CuO

then do the same for

H = 2.02/1

=2.02

so CuO is the limiting reagent because there is less amount of it.

Hope this helps  :)

4 0
2 years ago
A student fails to clean the pipet first. after delivering the vinegar sample, the student notices a drop of vinegar clinging to
VladimirAG [237]
This should not matter because the pipet has gradations and usually more of the sample is taken up in the pipette than what is delivered into the flask the student should always rinse the container being used because they are contaminating the sample if they do not clean it out
4 0
2 years ago
A balloon is filled with 0.400 moles of gas. How many liters of gas at STP were pumped into the balloon?
Fynjy0 [20]

Answer:

8.96 L

Explanation:

At STP, 1 mole = 22.4 L

0.400 mole *   (22.4 L. /1 mole of gas) = 8.96 L

3 0
3 years ago
A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
kondor19780726 [428]

Answer:

The final volume of the cylinder is 1.67 L

Explanation:

Step 1: Data given

Initial volume = 0.250 L

external pressure = 2.00 atm

Expansion does 288 J of work on the surroundings

Step 2: Definition of reversible work:

Wrev = -P(V2-V1) = -288 J

The gas did work, so V2>V1  (volume expands) and the work has a negative sign.(Wrev<0)

V2 = (-Wrev/P)  + V1

⇒ with Wrev = reverse work (in J)

⇒ with P = the external pressure (in atm)

⇒ with V1 = the initial volume

We can see that your pressure is in  atm  and energy in J

To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :

R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol

V2 =- (-288 J * (0.08206 L*atm/K*mol  /8.314 J/mol *K))/2.00 atm  + 0.250L

V2 = 1.67 L

The final volume of the cylinder is 1.67 L

8 0
3 years ago
What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

3 0
3 years ago
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