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3 years ago

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According to the early classification system plants were put in the following categories: 1.____________ 2.

____________ 3.______________________

1 answer:

rosijanka [135]3 years ago

3 0

We need more information to answer that

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The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

<u>Answer:</u> The moles of $CO_2$ added to the system is 7.13 moles

<u>Explanation:</u>

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

4 0

3 years ago

Considering the chart shown, at what temperature does the substance boil? 200ºC 1,200ºC 2,200ºC 2,400ºC

kodGreya [7K]

Answer:

The answer is 2,200ºC

Explanation:
I took the assignment for Edge, I don't think I can send the image because it might pick up on that and get reported, sorry!

The only other rationale that I have is that it's boiling because the graph shows that it's at a constant temperature/rate at 2,200ºC for quite a while. Typically when something boils, it stays at that constant rate of boiling, unless you turn the temperature up or it's finally able to peak..?

8 0

3 years ago

Sio2(s)+3c(s)→sic(s)+2co(g)

bonufazy [111]

Please explain in more detail

7 0

3 years ago

A woman with diagnosed osteoporosis is told by her doctor to take 1,000 mg of calcium carbonate in addition to increasing intake

spayn [35]

Answer:

400 mg

Explanation:

Given:

A woman with is told by her doctor to take 1,000 mg of calcium carbonate.

Question asked:

How many mg of calcium are in this 1,000 mg regimen of calcium carbonate?

Solution:

As we know:

<em>Amount of elemental calcium in a medicine containing calcium carbonate = 40%</em>

As here doctor prescribed 1000 mg of calcium carbonate to be taken by the woman increasing intake of calcium-containing foods in her diet and including regular weight-bearing exercise, hence amount of elemental calcium in 1,000 mg regimen of calcium carbonate will be,

$40\%\ of\ 1000\ mg\\\frac{40}{100} \times1000\ mg\\\frac{40000}{100} = 400\ mg$

Therefore, 400 mg of calcium are in this 1,000 mg regimen of calcium carbonate.

6 0

4 years ago

How does the number of protons compare to the number of<br> electrons in a canton

Tema [17]

Answer: The number of protons in cation is more than the number of electrons

Explanation:

cations are positive ions that means that they lose electrons to be stable nd have electron configuration (atomic number ) equivalent to the preceding noble gas

as : sodium 11 , it loses on electron to be stable and its electron configuration is equivalent to neon 10

so ,the number of protons in cation is more than the number of electrons

i hope it helps!

7 0

3 years ago