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3 years ago

Theories change because new technology improves our ability to observe the natural world.

Chemistry

1 answer:

rewona [7]3 years ago

7 0

Answer:

True

Theories change because new technology improves our ability to observe the natural world. answer choices.

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The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) CCl4(g) Calculate the equilibrium p

Ostrovityanka [42]

The complete question is as follows: The equilibrium constant, Kp, for the following reaction is 10.5 at 350 K: 2CH2Cl2(g) CH4(g) + CCl4(g) Calculate the equilibrium partial pressures of all species when CH2Cl2(g) is introduced into an evacuated flask at a pressure of 0.968 atm at 350 K. PCH2Cl2 = atm PCH4 = atm PCCl4 = atm

Answer: The equilibrium partial pressures of all species, that is, $CH_{4}$ , $CCl_{4}$ and $CH_{2}Cl_{2}$ is 0.420 atm, 0.420 atm and 0.128 atm.

Explanation:

For the given reaction equation, the initial and equilibrium concentration of involved species is as follows.

$2CH_{2}Cl_{2}(g) \rightarrow CH_{4}(g) + CCl_{4}(g)\\$

Initial: 0.968 atm 0 0

Equilibrium: (0.968 - 2x) x x

Now, $K_{p}$ for this reaction is as follows.

$K_{p} = \frac{P_{CH_{4}}P_{CCl_{4}}}{P^{2}_{CH_{2}Cl_{2}}}\\10.5 = \frac{x \times x}{(0.968 - 2x)^{2}}\\x = 0.420$

$P_{CH_{4}} = x = 0.420 atm\\P_{CCl_{4}} = x = 0.420 atm\\P_{CH_{2}Cl_{2}} = (0.968 atm - 2x) = (0.968 atm - 2(0.420)) = 0.128 atm$

Thus, we can conclude that the equilibrium partial pressures of all species, that is, $CH_{4}$ , $CCl_{4}$ and $CH_{2}Cl_{2}$ is 0.420 atm, 0.420 atm and 0.128 atm.

3 0

3 years ago

The surface area of an object to be gold plated is 49.8 cm2 and the density of gold is 19.3 g/cm-3. A current of 3.15. A is appl

garik1379 [7]

Answer: Time required to deposit an even layer of gold with given thickness is $5.3 \times 10^{2}$ sec.

Explanation:

The given data is as follows.

Surface area = 49.8 $cm^{2}$ ,

Density of gold = 19.3 $g/cm^{3}$ ,

Current = 3.15 A, thickness of gold layer = $1.2 \times 10^{-3} cm$

It is known that relation between volume, area and thickness is as follows.

V = Surface area × Thickness

= $49.8 \times 1.2 \times 10^{-3} cm$

= 0.05988 $cm^{3}$

Therefore, we will calculate the time required to deposit an even layer of gold with given thickness is calculated as follows.

$0.05988 \times cm^{3} \times \frac{19.3 g Au}{1 cm^{3}} \times \frac{1 mol Au}{197 g Au} \times \frac{3 mol e^{-}}{1 mol Au} \times \frac{96485}{1 mol e^{-}} \times \frac{1 As}{1 C} \times \frac{1}{3.20 A}$

= $5.3 \times 10^{2}$ sec

Thus, we can conclude that time required to deposit an even layer of gold with given thickness is $5.3 \times 10^{2}$ sec.

4 0

3 years ago

Whats 2 +5 + 6 + 9 right awnser please

masya89 [10]

Answer:

the awnser is 22

Explanation:

9+6 = 15 + 7 = 22 have a great day

5 0

2 years ago

Read 2 more answers

How does a planet’s motion affect its length of day and year?

rosijanka [135]

Answer:

Every location on Earth experiences an average of 12 hours of light per day but the actual number of hours of daylight on any particular day of the year varies from place to place.

5 0

3 years ago

How many moles of nitrogen dioxide are required to produce 3.56 moles of nitric acid?

alekssr [168]

Considering the reaction:
3NO₂ + H₂O → 2HNO₃ + NO

To produce 3.56 moles of nitric acid, we require:
3.56 x 3/2 moles of NO₂
Moles of NO₂ = 5.34 moles of NO₂

8 0

3 years ago