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exis [7]
3 years ago
12

Which of the following compounds undergoes

Chemistry
1 answer:
uysha [10]3 years ago
7 0

Answer:

A. C2H4

Ethene

step by step explanation:

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Part A
katovenus [111]

Answer:1.7

Explanation:

6 0
3 years ago
Why is the collision theory considered to be well-established and highly reliable?
hodyreva [135]

BECAUSE IT NEVER BREAKES


5 0
4 years ago
Objects in the universe vary in their composition. Which of the following best describes the composition of a galaxy?
professor190 [17]

Answer:

d

Explanation:

i looked it up

5 0
3 years ago
Read 2 more answers
Name the person who developed a table of elements which revealed
aivan3 [116]

Dimitry Mendeleef

Explanation:

Dimitry Mendeleef developed a table of elements which revealed regularities in elemental properties. He discovered the periodic arrangement of elements in 1869 while playing his game of solitaire.

  • John Newlands in in 1863 proposed the idea of a repeating octaves of properties of elements.
  • Dimitry Mendeleef described the periodic table or chart in 1869
  • The chart was based on the periodic law which states that "chemical properties of elements are a periodic function of their atomic weights".
  • In his periodic table, elements were arranged by atomic weights with recurring properties in a periodic manner>

Learn more:

Periodic table brainly.com/question/8543126

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8 0
3 years ago
A 10.0 L flask at 318 K contains a mixture of Ar and CH4 with a total pressure of 1.040 atm. If the mole fraction of Ar is 0.715
andrey2020 [161]

Answer:

w_{Ar}=0.814

Explanation:

Hello,

In this case, given the temperature, volume and total pressure, we can compute the total moles by using the ideal gas equation:

PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.040atm*10.0L}{0.082\frac{atm*L}{mol*K}*318K}\\  \\n=0.41mol

Next, using the molar fraction of argon, we compute the moles of argon:

n_{Ar}=0.41mol*0.715=0.29mol

And the moles of methane:

n_{CH_4}=0.41mol-0.29mol=0.12mol

Now, by using the molar masses of both argon and methane, we can compute the mass percent of argon:

w_{Ar}=\frac{m_{Ar}}{m_{Ar}+m_{CH_4}}=\frac{0.29mol*\frac{40g}{1mol}  }{0.29mol*\frac{40g}{1mol}+0.12mol*\frac{16g}{1mol}}  \\\\w_{Ar}=0.814

Regards.

7 0
3 years ago
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