Which statement describes continental crust

A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th

kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = $\frac{38.8}{61.2^{2} }$

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

0.0104 = $\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}$

0.0104 $[P(NO_{2} )]^{2}$ + $P(NO_{2} )$ - 200 = 0

Resolving the second degree equation:

$P(NO_{2} )$ = $\frac{-1+\sqrt{9.32} }{0.0208}$

$P(NO_{2} )$ = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0

3 years ago

How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potass

sergiy2304 [10]

The amount of silver chromate that precipitates after addition of solutions is 12.44 g.

Number of moles:

The number of moles is the product of molarity of the solution and its volume. The formula is expressed as:

Moles = Molarity x Volume

Calculations:

Step 1:

The molecular formula of silver nitrate is AgNO3. The number of moles of silver nitrate is calculated as:

Moles of AgNO3 = 0.500 M x (150/1000) L

= 0.075 mol

Step 2:

The molecular formula potassium chromate is K2CrO4. The number of moles of potassium chromate is calculated as:

Moles of K2CrO4 = 0.400 M x (100/1000) L

= 0.04 mol

Step 3:

The balanced chemical reaction between AgNO3 and K2CrO4 is:

2AgNO3 + K2CrO4 -----> Ag2CrO4 + 2KNO3

The required number of moles of K2CrO4 = 0.075 mol/2 = 0.0375 mol

The given number of moles of K2CrO4 (0.04 mol) is more than the required number of moles (0.0375 mol). Therefore, AgNO3 is the limiting reagent.

Step 4:

According to the reaction, the molar ratio between AgNO3 and Ag2CrO4 is 2:1. Hence, the number of moles of Ag2CrO4 formed is 0.0375 mol.

The molar mass of Ag2CrO4 is 331.74 g/mol.

The mass of Ag2CrO4 is calculated as:

Mass = 0.0375 mol x 331.74 g/mol

= 12.44 g

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6 0

2 years ago

Calculate the number of moles<br> 41.0 g of Ca(NO3)2 <br> 1.48 g of Ca(OH)2 <br> 3.40 g of CaSO4

Lemur [1.5K]

First. moles is just a label for a number of things. just like a dozen = 12, a gross = 144, a mole = 6022 with another 20 zeros after the 2

next

moles = mass / molecular weight.

molecular weight = sum of atomic mass from the periodic table

atomic mass MnO2 = atomic mass Mn + 2 x atomic mass O
= 54.94 + 2 x 16 = 86.94 g/mole

so moles MnO2 = 98.0 grams / (86.94 g/mole) = 1.13 moles

notice that I only gave 3 digits? that because of sig figs read the link below if you don't understand....

mw C5H12 = 5 x 12 + 12 x 1 = 72 g/mole

so moles C5H12 = 12.0 g / 72.0 g/mole = 0.167 moles

mw XeF6 = 131.3+ 6 x 19.00 = 245.3

so moles XeF6 = 100 g / 245.3 g/mole = 0.4077 moles

I've also provided a link to a periodic table. if you need atomic weights click on any element and it will give you the
details.

8 0

3 years ago

Element X has a charge of +2 and Element Y has a charge of -1.

bekas [8.4K]

the chemical equation will be XY2

6 0

3 years ago

5.00 moles of a binary, group 2 oxide are found to have a mass of 521 g. Identify the group 2 metal

andrey2020 [161]

5.00 moles of a binary, group 2 oxide are found to have a mass of 521 g. The group 2 metal is Strontium.

According to question 5 moles of binary group 2 metal oxide mass =521 g

So, 1-mole metal oxide mass will be =521 / 5=104.2 g

Now, as metal oxide is group 2 oxide so metal: oxygen =1: 1

So the mass of metal is =104.2-16=88.2 gm (as the atomic mass of oxygen is 16 g

Therefore, the metal is Strontium(Sr) whose atomic mass is 87.62 g which is nearly 88.2 gm

So, the formula of metal oxide is SrO

The group two metal here is Sr which is strontium and lies in the same group as calcium.

To learn more about group 2 metals, visit:

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4 0

2 years ago