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Margarita [4]
3 years ago
6

Down there (pls hurryy!!)

Chemistry
2 answers:
True [87]3 years ago
8 0

Answer:

where is the graph?

Explanation:

I cant answer it if there isnt a graph

Stella [2.4K]3 years ago
7 0

Answer:

250 kg answer

Explanation:

the activation energy for the reverse reaction is 250 kg and make a bar graph from 50 to 250 point it

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Scientific knowledge results from the interaction of scientists who are investigating natural phenomena. Which of these BEST des
seraphim [82]

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5ywhxhghghghfhffgffgffg

Explanation:

6 0
3 years ago
What purpose does ammonia lauryl sulfate serve as a surfactor
Alborosie

Answer:

See below  

Step-by-step explanation:

Ammonium lauryl sulfate has the structural formula CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂OSO₂O⁻ NH₄⁺.

The long nonpolar hydrocarbon chain and the ionic sulfate end group make it a surfactant.  

The ionic end tends to dissolve in water, but the nonpolar chain does not. This makes the compound an excellent <em>foaming agent,</em> so it is used in many shampoos and toothpastes.

The molecules form <em>micelles</em> in water, small spherical shapes with the polar heads outside, facing the water, and the nonpolar tails are inside.

They reduce the surface tension or the water so that, when you brush your teeth or shampoo your hair, the air bubbles are stable and do not break.

8 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
3 years ago
The process of chlorination in water treatment aims at ?​
Andrews [41]

Answer:

Chlorowanie wody jest procesem względnie tanim – koszt zbudowania samej instalacji do dezynfekcji, jak również koszt operacyjny wytworzenia 1 metra sześciennego uzdatnionej chlorem wody jest niższy od dezynfekcji wody za pomocą ozonu. Źródłem chloru w procesie dezynfekcji jest zazwyczaj podchloryn sodu, rzadziej dwutlenek chloru.

Explanation:

8 0
2 years ago
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
3 years ago
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