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3 years ago

Two examples of colloids

Chemistry

1 answer:

erica [24]3 years ago

8 0

${\huge {\mathfrak {\orange {Answer :-}}}}$

➡️ Milk

➡️ Butter

➡️ Muddy water

➡️ Smoke

REGARDS, ARNAB ✌️

THANK YOU

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How much energy is absorbed when 10 grams of gold increases in

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Answer:

6.45 joules

Explanation:

energy=10×(55-50)×0.129

=1.29×5

=6.45 J

hope this helps

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3 years ago

How can i prove that nascent oxygen more reactive than normal oxygen ?

laila [671]

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3 years ago

Indicate which of the following criteria are important for the selection of a buffer to use in an in vitro biochemical reaction.

JulsSmile [24]

Answer:

The correct answer is: d. The pKa of the chosen buffer should be close to the optimal pH for the biochemical reaction.

Explanation:

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3 years ago

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cricket20 [7]

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3 years ago

Read 2 more answers

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