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Sophie [7]
3 years ago
6

How does the average reaction rate differ from an instantaneous reaction rate?

Chemistry
1 answer:
____ [38]3 years ago
7 0

Answer:

The average rate is the change in concentration over a selected period of time. It depends on when you take the measurements. The instantaneous rate is the rate at a particular time.

Explanation:

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Dust particles in the air
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0.47 M KNO3 solution contains .757475 mol of KNO3 what is the volume
vfiekz [6]

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Volume is the quantity of three-dimensional  

Explanation:

7 0
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Which pair will for an ionic compound
vovikov84 [41]

Answer: A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.

Explanation:

A pair of elements will most likely form an ionic bond if one is a metal and one is a nonmetal. These types of ionic compounds are composed of monatomic cations and anions.

Explanation:

The chart below shows monatomic ions formed when an atom loses or gains one or more electrons, and the ionic compounds they form. You can check your periodic table to see that the cations are monatomic ions formed from metals, and the anions are monatomic ions formed from nonmetals.

3 0
2 years ago
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
torisob [31]

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

6 0
3 years ago
Do you think these fish are an example of natural selection among two populations of the same species
pshichka [43]
I don’t see any fish or anything
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3 years ago
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