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nataly862011 [7]
2 years ago
10

How many grams of KCl 03 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?

Chemistry
1 answer:
Nina [5.8K]2 years ago
3 0

11.48-gram of KCl0_3 are needed to produce 6.75 Liters of O_2  gas measured at 1.3 atm pressure and 298 K

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1.3 atm

V= 6.75 Liters

n=?

R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}

T=298 K

Putting value in the given equation:

\frac{PV}{RT}=n

n= \frac{1.3 \;atm\; X \;6.75 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 298}

Moles = 0.3588 moles

Now,

Moles = \frac{mass}{molar \;mass}

0.3588 moles = \frac{mass}{32}

Mass= 11.48 gram

Hence, 11.48-gram of KCl0_3 are needed to produce 6.75 Liters of O_2 gas measured at 1.3 atm pressure and 298 K

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

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if 28.5 g of calcium hydroxide is dissolved in enough water to make 185g of solution what is the percent by mass of calcium hydr
DENIUS [597]

The percent by mass of calcium hydroxide in the solution : 15.41%

<h3>Further explanation</h3>

The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight/volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.

Mass of solute (Ca(OH₂-Calcium hydroxide) : 28.5

Mass of solution = 185 g

\tt \%mass=\dfrac{mass~solute}{mass~solution}\times 100\%\\\\\%mass=\dfrac{28.5~g}{185~g}\times 100\%\\\\\%mass=15.41\%

6 0
3 years ago
Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
What is the concentration of a solution with a volume of 660L that contains 33.4g of AlCO3?
lana [24]

Answer:

The concentration of the solution is 5.8168 × 10^{-4} mol.dm^{-3}

Explanation:

Here, we want to calculate the concentration of the solution.

The unit of this is mol/dm^3

So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3

The number of moles = mass/molar mass

molar mass of AlCO3 = 27 + 12 + 3(16)  = 27 + 12 + 48 = 87g/mol

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This 0.384 moles is present in 660 L

x moles will be present in 1 dm^3

Recall 1 dm^3 = 1L

x * 660 = 0.384 * 1

x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3

7 0
2 years ago
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Licemer1 [7]

Answer: Option (3) is the correct answer.

Explanation:

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Atomic number of iron is 26. So, number of electrons present in Fe^{2+} is 26 - 2 = 24 electrons.

Atomic number of vanadium is 23. So, number of electrons present in V is 23 electrons.

Atomic number of scandium is 21. So, number of electrons present in Sc^{2-} is 21 + 2 = 23 electrons.

Thus, we can conclude that out of the given species, Fe^{2+} has the greatest  number of electrons.

4 0
3 years ago
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Answer:

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Hope this helps!

7 0
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