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2 years ago

Most reactions speed up when

2 answers:

7 0

For a reaction to occur, there should be mobility of ions in reactant side.
If the reactant is larger, its mobility will be lesser than that of smaller ones.
So reactants smaller in size have higher mobility which makes reaction faster.

Hence D is the correct option.

Hope this helps, have a great day/night ahead!

ehidna [41]2 years ago

6 0

The answer to this is D

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What is the mass of oxygen in 300 grams of carbonic acid (H2CO3)

DENIUS [597]

mass of carbonic acid = 300g

molar mass of H2CO3 = 2H + C + 3 O

= 2 x 1.008+ 12.01 + 3 x 16

= 62.03g/mol

moles of H2CO3 = mass/Molar mass

= 300/62.03

= 4.8364 moles

1 mole H2CO3 has 3 moles Oxygen

4.8364 moles H2CO3 contains

= 3 x 4.8364 moles Oxygen = 14.509 moles Oxygen

moles = mass/Molar mass

mass of oxygen = moles x Molar mass of Oxygen

= 14.509 x 16

= 232.15g Oxygen

mass of oxygen in 300g of carbonic acid(H2CO3) = 232.15g

7 0

2 years ago

Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance

ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0

2 years ago

Is air matter or energy

MatroZZZ [7]

Air is matter bc it is a gas

8 0

2 years ago

Read 2 more answers

Logging trees is often controversial. Why would some people support logging and some people not support logging?

Harlamova29_29 [7]

It seems more and more there are fewer conservation organizations who speak for the forest, and more that speak for the timber industry. Witness several recent commentaries in Oregon papers that are by no means unique. I’ve seen similar themes from other conservation groups across the West in recent years.

Many conservation groups have uncritically adopted views that support more logging of our public lands based upon increasingly disputed ideas about forest health and fire ecology, as well as the age-old bias against natural processes like wildfire and beetles.

For instance, an article in the Portland Oregonian quotes Oregon Wild’s executive director Sean Stevens bemoaning the closure of a timber mill in John Day Oregon. Stevens said: “Loss of the 29-year-old Malheur Lumber Co. mill would be ‘a sad turn of events’” Surprisingly, Oregon Wild is readily supporting federal subsidies to promote more logging on the Malheur National Forest to sustain the mill.

7 0

2 years ago

Read 2 more answers

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride

IrinaK [193]

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

<u>Answer:</u> The volume of hydrogen chloride produced in the reaction will be 4.4 mL

<u>Explanation:</u>

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

$CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)$

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$