Determine the rate of a reaction that follows the rate law:

A gas at 200 K occupies a volume of 350 ml. What temperature is needed to increase the volume

Rudik [331]

Answer: Temperature in constant pressure is 286 K

Explanation: If pressure remains constant, then V/T = constant.

V1 = 350 ml and T1 = 200 K and V2 = 500 ml. V1/T1 = V2/T2

and T2 = T1· V2 / V1 = 200 K · 500 ml / 300 ml = 285,7 K

5 0

3 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

By titration, it is found that 28.5 mL of 0.183 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the

maksim [4K]

Answer:

0.209M

Explanation:

M1V1=M2V2

(28.5 mL)(0.183M)=(25.0mL)(M)

M2= 0.209M

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4 0

3 years ago

The Lucas test has _______ results based on the type of alcohol present because the reaction involves a _________, which is ____

Olin [163]

Answer:

1) positive

2) carbocation

3) most stable

4) faster

Explanation:

A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.

The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.

Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.

Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.

5 0

3 years ago

The mole concept is important in chemistry because 1. it allows us to distinguish between elements and compounds. 2. it provides

Rufina [12.5K]

Answer:
<span>The mole concept is important in chemistry because, "</span>Atoms and molecules are very small and the mole concept allows us to count atoms and molecules by weighing macroscopic amounts of material".

Explanation:
To understand this question lets take an example of Hydrogen atom. Let suppose you need to react Hydrogen with Oxygen. You need exactly Two Hydrogen atoms and one Oxygen atom to form one water molecule.
The mass of 1 hydrogen atom is 1.76 × 10⁻²⁴ grams. How will you count the Hydrogen atoms??? How can you measure exactly for 1 Million Hydrogen Atoms???
Answer to these questions and Calculations lies in Mole. It is found that 1 Mole of Hydrogen weights exactly 1.008 gram and contains 6.022 × 10²³ atoms. Now, having this reference in hand you can calculate for any number of Hydrogen atoms.

Result:
So the Mole helps us to zoom a microscopic level to a macroscopic level. :)

6 0

3 years ago