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What ions are present when (nh4)2so4 is dissolved in water?

maksim [4K]

NH $NH_{4} ^{+}, SO _{4} ^{2-}$ are the two ions

8 0

3 years ago

Which of the following aqueous solutions will have the lowest freezing point? a. 0.75 m (NH4)3PO4 b. 1.0 m CaSO4 c. 1.0 m LiCl d

Nataly [62]

Answer: a. 0.75m (NH4)3PO4 will have the lowest freezing point .

Explanation: Freezing point decreases as the concentration of the solute substance increases. Assuming the same solvent for all of them, for instance water.

∆T= i.Kf.b

∆T= freezing point depression

i= vant Hoff factor

Kf= molality

Assuming water to be the solvent for all Kf=1.86°C/M

VANT HOFF FACTORS :

For (NH4)3PO4

This has 3 ionic bonding and 1 covalent bonding making it 4 bond

Therefore i=4

For CaSO4

This has 1 metallic bond and 1 covalent bond making it 2 bond.

Therefore i=2

For LiCl

This has 1 metallic bond and 1 non metallic bond making it 2 bond.

Therefore i=2

For CH3OH

This has only 1 covalent bond.

Therefore i=1

MOLALITY:

(NH4)3PO4 = 0.75M

CaSO4= 1.0M

LiCl= 1.0M

CH3OH= 1.5M

FREEZING POINT DEPRESSION:

For (NH4)3PO4

∆T= 4×0.75×1.86=5.58°C

For CaSO4

∆T= 2×1.0×1.86=3.72°C

For LiCl

∆T = 2×1.0×1.86= 3.72°C

For CH3OH

∆T= 1×1.5×1.86=2.79°C

REMEMBER THE HIGHER THE FREEZING POINT DEPRESSION THE LOWER THE FREEZING POINT.

FREEZING POINT DEPRESSION IS THE CHANGE IN THE FREEZING POINT PROPORTIONAL TO THE AMOUNT OF SOLUTE ADDED THE THE SOLUTION.

THEREFORE THE ONE WITH THE LOWEST FREEZING POINT IS (NH4)3PO4

4 0

4 years ago

Which statement best summarizes what happens during cellular respiration?

morpeh [17]

Answer:

A

Explanation:

B and D have to do with plants

and C has nothing to do with cellular resporation.

8 0

3 years ago

Read 2 more answers

Oxygen 16 has 8 protons how many neutrons

KIM [24]

There are 8 neutrons. The mass number subtracted by the atomic number gives you neutrons. if your continuing with isotopes, find the average mass of Oxygen 16

5 0

4 years ago

Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) â†’ 2Al2O3 (s) In

jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

$\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol$

For oxygen gas:

$\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol$

The chemical equation for the reaction of titanium and chlorine gas follows:

$4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)$

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = $\frac{3}{4}\times 0.092=0.069mol$ of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of $Al_2O_3$

So, 0.092 moles of aluminium will produce = $\frac{2}{4}\times 0.092=0.046moles$ of $Al_2O_3$

Now, calculating the mass of aluminium oxide:

$\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g$

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

$\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%$

Hence, the percent yield of the reaction is 74 %

6 0

3 years ago