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drek231 [11]
3 years ago
10

An idea is being proposed. The steps that led to the idea are listed below:

Chemistry
1 answer:
Lena [83]3 years ago
3 0

Answer:

No because opinion and social values may lead to bias

Explanation:

I'm smart >:)

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kakasveta [241]
No entiendo cual es tu pregunta?
3 0
3 years ago
After the first steps in writing an equation, the equation is balanced by
kifflom [539]

Answer:

law of conservation of mass

Explanation:

self explanatory

5 0
3 years ago
Can someone help me with this I've been stuck on it for a few days​
Fiesta28 [93]
So I haven’t got time to answer all of it for you but the id you look at the picture of the periodic table I’ve added the top number in the red boxes are the groups and the period is how many elements down from the top it is (remember that the hydrogen and helium make up period ONE) so remember to include them when counting the elements as you go down the table

5 0
2 years ago
Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the sys
8090 [49]

Answer:

Choice d. No effect will be observed as long as other factors (temperature, in particular) are unchanged.

Explanation:

The equilibrium constant of a reaction does not depend on the pressure. For this particular reaction, the equilibrium quotient is:

Q = \displaystyle \frac{[\mathrm{SO_2\, (g)}]}{[\mathrm{O_2\, (g)}]}.

Note that the two sides of this balanced equation contain an equal number of gaseous particles. Indeed, both [\mathrm{SO_2\, (g)}] and [\mathrm{O_2\, (g)}] will increase if the pressure is increased through compression. However, because \rm SO_2\, (g) and \rm O_2\, (g) have the same coefficients in the equation, their concentrations are raised to the same power in the equilibrium quotient Q.

As a result, the increase in pressure will have no impact on the value of Q\!. If the system was already at equilibrium, it will continue to be at an equilibrium even after the change to its pressure. Therefore, no overall effect on the equilibrium position should be visible.

8 0
3 years ago
Hi guys, my question is:
guapka [62]

Answer:

See Explanation Below

Explanation:

A) The rate law can only be on the reactant side and you can only determine it after you get the net ionic equation because of spectators cancelling out. So in this case the rate law is k=[CH3Br]^1 [OH-]^1. The powers are there because the rxn is first order.  

B) Since the rxn is first order anything you do to it will be the exact same "counter rxn" per say so since you are decreasing the OH- by 5 the rate will decease by 5

C) The rate will increase by 4 since you are doubling both you have to multiply them both.

8 0
3 years ago
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