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Sholpan [36]
3 years ago
10

I need help on this :/

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer: b

Explanation:

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A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.
Lilit [14]

Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

               \frac{V1}{T1} = \frac{V2}{T2}

Solve for V2

                V2 = \frac{V1T2}{T1}

3.- Substitution

               V2 = \frac{(42)(233)}{293}

4.- Simplification

                V2 = \frac{9786}{293}

5.- Result

                V2 = 33.4ml

3 0
3 years ago
What is the similarity between radioactive iodine and stable iodine
Elena-2011 [213]

Answer:

both iodine

Explanation:

7 0
4 years ago
Write the balanced equation for the formation of carbon dioxide.
ohaa [14]
C + O2 ——> CO2
Carbon plus oxygen forms carbon dioxide
4 0
3 years ago
What is the IUPAC name of this compound?  ________ CH3-CHCl-CH2-CH2-Cl​
Diano4ka-milaya [45]

Answer:

The prefixes are fluoro-, chloro-, bromo-, and iodo-. Thus CH 3CH 2Cl has the common name ethyl chloride and the IUPAC name chloroethane. Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names.05‏/06‏/2019

6 0
3 years ago
Read 2 more answers
Ethylene (C2H4) is the starting material for the preparation of polyethylene. Although typically made during the processing of p
Solnce55 [7]

Answer:

1.17 grams

Explanation:

Let's consider the balanced equation for the combustion of ethylene.

C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)

We can establish the following relations:

  • 1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
  • The molar mass of C₂H₄ is 28.05 g/mol.

The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}

8 0
3 years ago
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