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vredina [299]
1 year ago
11

Calculate the number of molecules in 1500 mL of a gas measured under a pressure of 800.0

Chemistry
1 answer:
k0ka [10]1 year ago
5 0

Answer:

3.97 x 10²² molecules

Explanation:

To find the moles, you need to use the Ideal Gas Law:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = Ideal Gas constant (0.08206 atm*L/mol*K)

-----> T = temperature (K)

Before you can plug the values into the equation, you need to (1) convert the pressure from torr to atm (by dividing by 760), then (2) convert the volume from mL to L (by dividing by 1000), and then (3) convert the temperature from Celsius to Kelvin (by adding 273).

P = 800.0 torr / 760 = 1.05 atm               R = 0.08206 atm*L/mol*K

V = 1500 mL / 1000 = 1.5 L                       T = 19.0 °C + 273 = 292 K

n = ? moles

PV = nRT

(1.05 atm)(1.5 L) = n(0.08206 atm*L/mol*K)(292 K)

1.579 = n(23.96)

0.0659 = moles

Now, you need to convert moles to molecules using Avogadro's Number.

Avogadro's Number:

6.022 x 10²³ molecules = 1 mole

 0.0659 moles          6.022 x 10²³ molecules
------------------------  x  -------------------------------------  =  3.97 x 10²² molecules
                                               1 mole

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yuradex [85]

Answer:

2 Al and 3 S

Explanation:

Al2S3 so 2 Al and 3 S

3 0
1 year ago
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
Consider the temperature versus time graph below. I which region is the substance in both the solid phase and the liquid phase?
SVEN [57.7K]

Answer:

2

Explanation:

The change of state occurs at a constant temperture and pressure. In the grahp we can see while the time passes, the temperature doesn't change.

The rect number 4 correspond to a liquid-gas phase

8 0
3 years ago
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RSB [31]
Big grasshoppers, because it does not involve a number.

Hope this helps! <3
7 0
3 years ago
If I have a 50.0 liter container that holds 45 moles of gas at a
Darina [25.2K]
V= 50. L n=45 mol T= 200°C = 473k P=?

CP)X 50.L)= (45 mol)(0.0821 light_kimol)(473k)


P = 30am or 4000 kPa
5 0
3 years ago
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