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3 years ago

Which techniques would be best for separating the solute from the solvent in a solution

Chemistry

1 answer:

grin007 [14]3 years ago

3 0

Answer:

depending on the solvent, evaporation would be the easiest

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Which of these is a covalent compound? A. LiCl B. MgO C. AlCl3 D. CO

Fiesta28 [93]

Answer:

C. AlCl3 is a covalent compound.

Explanation:

It is formed by mutual sharing of electrons between two atoms each contributing equal number of electrons to the electron pair.

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3 years ago

How many individual orbitals does carbon use for all of its electrons? Remember, there are two electrons in each filled orbital

Andreas93 [3]

Answer:

Three orbitals

Explanation:

The electronic configuration of carbon is given as follows;

1s²2s²2p²

Therefore, out of the six electrons of the carbon atoms, 4 fill the 1s and 2s orbitals with 2 electrons each, while the two remaining electrons are situated in the 2p orbital, with the electrons in the 2p orbital will remain unpaired such that they will have similar quantum numbers in accordance with Pauli exclusion principle.

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3 years ago

Which of the following statements is true about the composition of the atmosphere?

Phoenix [80]

Answer : Option D) The atmospheric conditions vary as one changes latitude and altitude.

Explanation : The composition of the atmosphere varies according to the latitude and altitude because of the unequal heating of the earth surface at different latitudes and altitudes which results into atmospheric changes. It also creates different regions and zones.

3 0

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 $NH_3$