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Wittaler [7]
2 years ago
11

Which plastic do you think could best contain extremely corrosive material? PETE, LDPE, PVC, PS, PP, or HDPE.

Chemistry
1 answer:
Viktor [21]2 years ago
7 0
The ones that are known to be corrosive are PETE and PVC. These plastics you need to avoid because these are <span>plastics that can be hydrolyzed by alkaline solutions. Hope this helps on your task</span>
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When 2.5 mol of O2 are consumed in their reaction, ________ mol of CO2 are produced
Vika [28.1K]

The given question is incomplete. The complete question is:

The combustion of propane (C3H8) in the presence of excess oxygen yields CO_2 and H_2O

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

When only 2.5 mol of O_2 are consumed in order to complete the reaction, ________ mol of CO_2 are produced.

Answer: Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

Explanation:

The balanced chemical equation is:

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2(g)+4H_2O (g)

According to stoichiometry :

5 moles of O_2 produce = 3 moles of  CO_2

Thus 2.5 moles of O_2 will produce = \frac{3}{5}\times 2.5=1.5 moles of  CO_2

Thus when 2.5 mol of O_2 are consumed in their reaction,  1.5 mol of CO_2 are produced

4 0
2 years ago
A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new
Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

PV=nRT

P= Pressure of the gas = 697 mmHg = 0.92 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

PV=nRT

P= Pressure of the gas = 775 mmHg = 1.02 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles

Thus the now the container contains 2.07 moles.

6 0
2 years ago
Imagine you needed to identify if an object has undergone a physical change or a chemical change. What information would you nee
pishuonlain [190]
How it looks. basically the thing that tells you how it change. for example if an ice cube was melted (heat), it only changed physically not chemically as the h20 molecules are still there. however lets say you burn woos— you cant get that would back. its ash now and it has changed chemically.
7 0
2 years ago
Of the following, ________ should have the highest critical temperature. Of the following, ________ should have the highest crit
3241004551 [841]

Answer:

H2

Explanation:

Critical temperature is the temperature above which gas cannot be liquefied, regardless of the pressure applied.

Critical temperature directly depends on the force of attraction between atoms, it means stronger the force of higher will be the critical temperature. So, from the given options H2 should have the highest critical temperature because of high attractive forces due to H bonding.

Hence, the correct option is H2.

5 0
2 years ago
In the peptide GLVIW, the C-terminal end is ________.
zheka24 [161]
The peptide given above is made up of the following amino acids: glycine [G], leucine [L], valine [V], isoleucine [I] and tryptophan [W]. These amino acids are joined together by amide bond to form peptide. Peptides usually have two terminals, the N terminal and the C terminal. For GLVIW, the C terminal end amino acid is tryptophan, that is the last amino acid on the peptide chain. The N terminal amino acid is glycine, that is, the first amino acid on the peptide chain.
5 0
2 years ago
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