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3 years ago

Convert 12.64 g of NaOH to moles

2 answers:

8 0

.316 moles just divide by the molar mass of naoh which is 39.997. So 12.64 divived by 39.997 is .3160237. that is 3.16x10-1

Sati [7]3 years ago

7 0

Molar mass of NaOH =22,990g/mol+15,999g/mol+1,001g/mol =39,99g/mol

n= mass/molar mass

moles of NaOH= 12,64 g * 1 mol NaOH /39,99g = 0,316 moles of NaOH in 12,64 g of NaOH

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Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon d

Korolek [52]

Answer:The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

Explanation:

Let the molecular formula of the oxide of metal be $X_2O_y$

$X_2O_y+yCO\rightarrrow yCO_2+2X$

Mass of metal product = 1.68 g

Moles of metal X = $\frac{1.68 g}{55.9 g/mol}=0.03005 mol$

1 mol of metal oxide produces 2 moles of metal X.

Then 0.03005 moles of metal X will be produced by:

$\frac{1}{2}\times 0.03005 mol=0.01502 mol$ of metal oxide

Mass of 0.01502 mol of metal oxide = 2.40 g (given)

$0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g$

y = 2.999 ≈ 3

The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

8 0

3 years ago

The solubility of glucose at 30°C is

weqwewe [10]

Answer:

Saturated solution

We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.

Explanation:

Step 1: Calculate the mass of water

The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.

$400 mL \times \frac{0.996g}{1mL} =398g$

Step 2: Calculate the mass of glucose per 100 g of water

550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.

$100gH_2O \times \frac{550gGlucose}{398gH_2O} = 138 gGlucose$

Step 3: Classify the solution

The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.

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4 years ago

How many moles of CO2 would be present in a gas sample of 10 L at 25.0oC and a pressure of .77 atm?

Step2247 [10]

Use PV=nRT to solve the equation. You need to solve for n (number of moles). Don’t forget to convert the temperature to kelvins by adding 25+273. Use 0.082057 for R.

4 0

3 years ago

How many moles of nitrogen are present at STP if the volume is 846L

Artyom0805 [142]

A mole of any gas occupied 22.4 L at STP. So, the number of moles of nitrogen gas at STP in 846 L would be 846/22.4 = 37.8 moles of nitrogen gas.

Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.

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Ipatiy [6.2K]

Answer:

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