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Oduvanchick [21]
3 years ago
10

How many liters of 1.2M solution can be prepared with 0.50 moles of C6H12O6

Chemistry
1 answer:
BigorU [14]3 years ago
5 0
First you need to know the molecular weight of sugar (C6H12O6) which is 180.156g/mol
You have half a mole so you have 90.078g
If you wanted to make 1L of a 1.2M solution of glucose you would need 180.156*1.2=216.1872g
But you only have 90.078g
So you need to figure out how much this 90.078g will make if the solution must be 1.2M:
90.078g/216.1872g=xL/1L
solve for the X and you get 0.416666666...
so 416.7ml or 0.417L
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Answer: product

Explanation:

Each substance written to the right of the arrow in a chemical equation is referred to as a product.

When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.

On the other hand which is the right side is the product.

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2 years ago
You have dissolved 10 g sodium oxide in 200 ml water.calculate concentration of the solution
e-lub [12.9K]

Answer:

0.85 Molar Na2O

Explanation:

Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).

 (10 g Na2O)/(61.98 g/mole) = 0.161 moles Na2O.

Molar is a measure of concentration.  It is defined as moles/liter.  A 1 M  solution contains 1 mole of solute per liter of solvent.  [200 ml water = 0.2 Liters water.]

In this case, we have 0.161 moles Na2O in 0.200 L of solvent.

 (0.161 moles Na2O)/(0.200 L) = 0.85 Molar Na2O

8 0
1 year ago
How many molecules of O2 are required to react with 3.00 moles P4?<br> P4+5O2⟶2P2O5
likoan [24]

Answer:

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8 0
3 years ago
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A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th
tatyana61 [14]

Answer : The partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % CO_2 and 25 % N_2 in terms of volume.

First we have to calculate the moles of CO_2 and N_2.

\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole

\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole

Now we have to calculate the mole fraction of CO_2.

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}

\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75

Now we have to calculate the partial pressure of the CO_2 gas.

\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}

\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the CO_2 in the tank in psia is, 32.6 psia.

3 0
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