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3 years ago

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Draw the structure for each compound below:________. A) Propanedial B) 4-Phenylbutanal C) (S )-3-Phenylbutanal D) 3,3,5,5-Tetram

ethyl-4-heptanone E) (R )-3-Hydroxypentanal F) meta-Hydroxyacetophenone G) 2,4,6-Trinitrobenzaldehyde H) Tribromoacetaldehyde I) (3R,4R )-3,4-Dihydroxy-2-pentanone

1 answer:

djverab [1.8K]3 years ago

7 0

Answer:

Please check attachment

Explanation:

In this question, we are to draw the structure for each of the compounds below.

Please check attachment for the diagrams

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The pressure of a fixed mass of gas is increased from 100 kPa to 600 kPa at a constant temperature. The new volume of the gas is

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Volume and pressure are inversely proportional (as one goes up, one goes down). So as you increase the pressure, you decrease the volume.

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What are the similarities between independent variable and a dependent variable?

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4 years ago

In a combustion experiment, it was found that 12.096 g of hydrogen molecules combined with 96.000 g of oxygen molecules to form

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The mass change, or the mass defect, can be calculated by the formula that is very known to be associated with Albert Einstein.

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3 years ago

Which mixture is most likely to form a suspension?flour and liquid water, as in gravy metal coins and liquid water sodium chlori

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3 years ago

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

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3 years ago