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Flauer [41]
3 years ago
5

PLEASE ANSWER QUICKLY

Chemistry
2 answers:
Margaret [11]3 years ago
3 0

Answer:  I say its Segmented body  and radial symmetry im not sure if it is right

Explanation:

ddd [48]3 years ago
3 0

Answer:

B. Nervous system

D. Segmented body

E. Open circulatory system

Explanation:

Hope this helps!

Have a nice dayy! :)

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Steel is an alloy that is mainly comprised of CARBON. Hope this helps and give thanks plsss!
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2.<br> C₂H6<br> +<br> O2<br> H20<br> +<br> CO2
kolbaska11 [484]

Answer:

C_{2}H_{6} +O_{2} -> H_{2}O+CO_2 is a combustion reaction

Explanation:

The reaction is a combustion since you have a hydrocarbon reacting with oxygen to form water and carbon dioxide.

The balanced equation is: 2C_{2}H_{6} +7O_{2} ->4H_{2}O+6CO_2

If this helped, a brainliest answer would be greatly appreciated!

4 0
3 years ago
If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500
ratelena [41]

<u>Answer:</u> The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

<u>Explanation:</u>

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

                         SO_2+NO_2\rightleftharpoons SO_3+NO

<u>Initial:</u>             0.500  0.500      0.005   0.005

<u>At eqllm:</u>      0.500-x  0.500-x   0.005+x  0.005+x

The expression of equilibrium constant for the above reaction follows:

K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}

We are given:

K_c=2.50

Putting values in above equation, we get:

2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = (0.500-x)=(0.500-0.304)=0.196M

Equilibrium concentration of nitrogen dioxide = (0.500-x)=(0.500-0.304)=0.196M

Equilibrium concentration of sulfur trioxide = (0.00500+x)=(0.00500+0.304)=0.309M

Equilibrium concentration of nitrogen monoxide = (0.00500+x)=(0.00500+0.304)=0.309M

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

4 0
3 years ago
Consider the following reversible reaction.
dmitriy555 [2]

Answer:

Keq = [CO₂]/[O₂]

Explanation:

Step 1: Write the balanced equation for the reaction at equilibrium

C(s) + O₂(g) ⇄ CO₂(g)

Step 2: Write the expression for the equilibrium constant (Keq)

The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:

Keq = [CO₂]/[O₂]

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The origin is what makes it change
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