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2 years ago

10

Which part of an energy graph is represented by the highest point?

2 answers:

Lilit [14]2 years ago

8 0

Provide an image or something

Ganezh [65]2 years ago

8 0

Answer:

The activated state

Explanation:

The activated state is the "highest energy of the chemicals as they proceed through a reaction". Therefore, they would be the highest point on the graph, since they are the highest energy level. The activated state can also represent the products or the reactants, dependent on the type of reaction. Hope this helps!

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Listed below are sets of elements.

Anna11 [10]

The correct answer is B) Chlorine, Sulfur, and Silicon

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3 years ago

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A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

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3 years ago

Why do we have to add acid to water but not water to acid

KATRIN_1 [288]

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3 years ago

Jack walks outside in the afternoon and noticed the clouds are very dark. He states that it looks like it will rain very soon. H

Kay [80]

Answer:

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3 years ago

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Carbon dioxide and water react to form methanol and oxygen, like this:

fredd [130]

Answer:

24x10³

Explanation:

2CO₂(g) + 4H₂O(g) → 2CH₃OH(l) + 3O₂ (g)

The equilibrium constant for this reaction is:

Kc = $\frac{[O_2]^3}{[CO_2]^2[H_2O]^4}$

The expression of [CH₃OH] is left out as it is a pure liquid.

Now we <u>convert the given masses of the relevant species into moles</u>, using their <em>respective molar masses</em>:

CO₂ ⇒ 3.28 g ÷ 44 g/mol = 0.0745 mol CO₂

H₂O ⇒ 3.86 g ÷ 18 g/mol = 0.214 mol H₂O

O₂ ⇒ 2.80 g ÷ 32 g/mol = 0.0875 mol O₂

Then we calculate the concentrations:

[CO₂] = 0.0745 mol / 7.5 L = 0.0099 M

[H₂O] = 0.214 mol / 7.5 L = 0.0285 M

[O₂] = 0.0875 mol / 7.5 L = 0.0117 M

Finally we <u>calculate Kc</u>:

Kc = $\frac{0.0117^3}{0.0099^2*0.0285^4}$ = 24x10³

3 0

3 years ago