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3 years ago

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200 grams of an organic sample which contains only carbon, hydrogen and oxygen is analyzed and found to contain 97.30grams of ca

rbon, 16.22grams of hydrogen and the remainder oxygen. what is the empirical formula for the compound?

1 answer:

Klio2033 [76]3 years ago

7 0

The empirical formula is C2H4O

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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

kodGreya [7K]

<u>Answer:</u> The net chemical equation for the production of acrylic acid is given below.

<u>Explanation:</u>

We are given two intermediate equations:

<u>Equation 1:</u> $CaC_2(s)+2H_2O(g)\rightarrow C_2H_2(g)+Ca(OH)_2(s)$

<u>Equation 2:</u> $6C_2H_2(g)+3CO_2(g)+4H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

To get the net chemical equation for the formation of acrylic acid from calcium carbide, carbon dioxide and water, we multiply the first equation by a factor of '6'.

The equation becomes:

<u>Equation 1:</u> $6C_2H_2(g)+6Ca(OH)_2(s)$

<u>Equation 2:</u> $6C_2H_2(g)+3CO_2(g)+4H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

Net chemical equation now becomes:

$6CaC_2(s)+16H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+Ca(OH)_2(s)$

Hence, the net chemical equation for the production of acrylic acid is given above.

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3 years ago

What are the twa parts of central nervous system ?

natali 33 [55]

The brain and the spinal cord

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3 years ago

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At 40°C, how much KNO3 is<br> soluble?

Ludmilka [50]

Answer: 63 / 100 g water

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4 years ago

What is the pOH of a<br> 2.6 x 10-6 M H+ solution?

melomori [17]

Answer:

Approximately $8.41$ (assuming that the solution is at $\rm 25^\circ C$ , under which $K_{\rm w} = 10^{-14}$ .)

Explanation:

Let ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ denote the concentration of $\rm H^{+}$ and $\rm OH^{-}$ respectively.

Let $K_{\rm w}$ denote the self-ionization constant of water. The exact value of $K_{\rm w}\!$ depends on the temperature of the solution. $K_{\rm w} =10^{-14}$ at $\rm 25^\circ C$ .

The product of ${\rm [H^{+}]}$ and ${\rm [OH^{-}]}$ in a solution (with $\rm M$ , or moles per liter, as the unit) is supposed to be equal to the $K_{\rm w}$ value of that solution at the corresponding temperature. In other words:

${\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}$ .

Rearrange to obtain an expression for ${[\rm OH^{-}]}$ :

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}$ .

Assume that the solution in this question is at $\rm 25^\circ C$ (for which $K_{\rm w} =10^{-14}$ .) For this solution:

$\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}$ .

Hence, the $\rm pOH$ of this solution would be:

$\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}$ .

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3 years ago

Which of the following is not a product of cellular respiration?

tatyana61 [14]

C is not a product of cellular respiration because C6H1206 is glucose and it is not a product of cellular respiration.

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3 years ago

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