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ArbitrLikvidat [17]
3 years ago
12

Which of the following statements is related to a postulate (assumption) of the kinetic molecular theory of gases?

Chemistry
1 answer:
Margaret [11]3 years ago
4 0
Gas particles are in constant motion and are very far apart compared to their size.
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I really need help solving this question
olchik [2.2K]
The answer to this would be helium
4 0
3 years ago
Read 2 more answers
Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.
Ronch [10]

Answer : The limiting reagent is O_2

Solution : Given,

Moles of methane = 2.8 moles

Moles of O_2 = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 5 moles of O_2 react with \frac{5}{2}=2.5 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is O_2

8 0
3 years ago
I AM GIVING A LOT OF POINTS SO PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
prohojiy [21]

Answer:c

Explanation:

it’s gained kinetic from the gravitational potential energy at the top

4 0
3 years ago
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The table above gives examples of inherited traits. Which statement below will BEST complete this table?
alexandr402 [8]
The answer would be a bird eats seeds because that is what the parent has been feeding him.
8 0
3 years ago
Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal
german

Answer:

The molecular formula of an ascenapthalene is C_{12}H_{10}

Explanation:

\Delta T_b=K_b\times m

\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}

where,

\Delta T_f =Elevation in boiling point = (62.5-61.7)^oC=0.8^oC

Mass of acenapthalene = 0.515 g

Mass of CHCl_3 = 15.0 g = 0.015 kg (1 kg = 1000 g)

K_b = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}

\text{Molar mass of acenapthalene}=155.7875 g/mol

Let the molecule formula of the Acenapthalene be C_{6n]H_{5n}

6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol

n = 2.0

The molecular formula of an ascenapthalene is C_{12}H_{10}

4 0
3 years ago
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