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Troyanec [42]
2 years ago
5

Identify the products and the reactants in the following chemical equations

Chemistry
1 answer:
liq [111]2 years ago
6 0

Answer:

1-Mg+O2=reactants                  MgO-product

2-CaNr2+AlCl3=reactants        CaCI2+AIBr3=product

3-C2H4+O2=reactants              H2O+CO2=product

Explanation:

Reactants are on the left side and products on the right.

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The answer is 0.08 mol
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Electrons in conductors, like copper wire, are free to roam throughout the metal in the presence of the ions that have released
Scrat [10]

Answer:

  • <u><em>No, I would not consider a metal to be a plasma because plasma is just another state of matter, and the copper wire is in solid state.</em></u>

Explanation:

Metal is not a state of matter. Metals can be solid or liquid (molten) depending on their melting point and the temperature at which they are.

Plasma is a state of matter, similar to gas, but it is reached only at very high temperatures like in the Sun. The particles in plasma state are not neutral atoms or molecules but negatively charged  ions and electrons.

The copper wire is yet a solid, thus it cannot be considered a plasma.

Metals can be in plasma state only if the temperature is too high, like the temperatures in the stars. In fact, the metals in the Sun and other hotter stars are in plasma state.

5 0
3 years ago
Which compound contains both sigma and pi bonds
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Any compound with multiple covalent bonds
6 0
3 years ago
Read 2 more answers
How many molecules are in 0.500 mole of N2O5?
Oksi-84 [34.3K]

Answer:

3,011.10e23.

Explanation:

3 0
3 years ago
If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reacti
ddd [48]

Answer:

Limiting reagent: barium nitrate

Theoretical yield: 2.29 g BaS

Percent yield: 87%

Explanation:

The corrected balanced reaction equation is:

K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS

The amount of potassium sulfide added is:

(25.0 mL)(1.20mol/L) = 30 mmol

The amount of barium nitrate added is:

(15.0mL)(0.900mol/L = 13.5 mmol

Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:

(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS

Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).

(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS

The percent yield is calculated as follows:

(actual yield) / (theoretical yield) x 100%

(2.0 g) / (2.29 g) x 100% = 87%

6 0
3 years ago
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