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tatuchka [14]
3 years ago
9

Just need the two blanks filled :)

Chemistry
1 answer:
Anestetic [448]3 years ago
5 0

Answer:

Dont answer this question its an online exam question

Explanation:

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While investigating greenhouse gases, I measured 0.1875 grams of carbon dioxide in a 500-gram sample of the atmosphere. What is
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[CO₂] = 375 ppm

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We can determine the concentration in value of ppm

ppm = mass of solute (mg) / mass of solution (kg)

Solution: atmosphere

Solute: CO₂

We convert the mass of CO₂ from g to mg → 0.1875 g . 1000 mg / 1g = 187.5 mg

We convert the mass of the atmosphere from g to kg: 500 g . 1 kg/1000g = 0.500 kg

ppm = 187.5 mg / 0.500kg = 375

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What is the coordination number of an iridium atom in the face-centered cubic structure of iridium?
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You can tell if an object is in motion if ___.
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3 years ago
Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas
hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Suppose 1.20 mol CS_2(g) of and 3.60 mol of Cl_2(g)  were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol  of CCl_4. Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of  CS_2 = 1.20 mole

Moles of  Cl_2 = 3.60 mole

Volume of solution = 1.00  L

Initial concentration of CS_2 = \frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M

Initial concentration of Cl_2 = \frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M

The given balanced equilibrium reaction is,

                 CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)

Initial conc.         1.20 M        3.60 M                  0                  0

At eqm. conc.     (1.20-x) M   (3.60-3x) M   (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}

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

K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}

Given :Equilibrium concentration of CCl_4 , x = \frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M

K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}

K_c=0.36

Thus equilibrium constant at unknown temperature is 0.36

4 0
3 years ago
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