How many grams of O are in 675 g of Na2O

Consider the following reaction:

Kitty [74]

Answer : The value of equilibrium constant (Kc) is, 0.0154

Explanation :

The given chemical reaction is:

$SO_2Cl_2(g)\rightarrow SO_2(g)+Cl_2(g)$

Initial conc. $2.4\times 10^{-2}$ 0 0

At eqm. $(2.4\times 10^{-2}-x)$ x x

As we are given:

Concentration of $Cl_2$ at equilibrium = $1.3\times 10^{-2}M$

That means,

$x=1.3\times 10^{-2}M$

The expression for equilibrium constant is:

$K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}$

Now put all the given values in this expression, we get:

$K_c=\frac{(x)\times (x)}{2.4\times 10^{-2}-x}$

$K_c=\frac{(1.3\times 10^{-2})\times (1.3\times 10^{-2})}{2.4\times 10^{-2}-1.3\times 10^{-2}}$

$K_c=0.0154$

Thus, the value of equilibrium constant (Kc) is, 0.0154

4 0

2 years ago

An

Basile [38]

Answer:

The answer of this question is molecule

4 0

2 years ago

The mass of a 50-mL volumetric flask is 40.638 g. After adding 15.0 mL of 95% ethanol and adding enough water to complete the vo

Nuetrik [128]

Answer

0.9516 grams / mL (50.00 has 4 sig digs.)

Remark

You have a couple of extraneous numbers there. You don't care about anything except the mass of the flask + water/alcohol mixture (88.219 grams). and the mass of the flask (40.638 grams)

Formulas

mass water/alcohol mixture = mass of the flask with fluid - mass flask
density = mass / volume

Solution

mass water/alcohol mixture = 88.219 - 40.638 = 47.581

Volume = 50 mL
Density = mass / Volume
Density = 47.581/50
Density = 0.95162 There are 4 sig digs so the answer should be
0.9516

3 0

3 years ago

Please please help me please please help please please help me please please help please please

Tatiana [17]

Answer:

AU9NJ-BLVHV-TCLJS-54YTB

6 0

2 years ago

At a certain temperature the vapor pressure of pure thiophene is measured to be . Suppose a solution is prepared by mixing of th

Lesechka [4]

Answer:

0.35 atm

Explanation:

It seems the question is incomplete. But an internet search shows me these values for the question:

" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."

Keep in mind that if the values in your question are different, your answer will be different too. However the methodology will remain the same.

First we calculate the moles of thiophene and heptane, using their molar mass:

137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene

111 g heptane ÷ 100 g/mol = 1.11 moles heptane

Total number of moles = 1.63 + 1.11 = 2.74 moles

The mole fraction of thiophene is:

1.63 / 2.74 = 0.59

Finally, the partial pressure of thiophene vapor is:

Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene

Partial Pressure = 0.59 * 0.60 atm

Pp = 0.35 atm

3 0

2 years ago