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love history [14]
4 years ago
11

Which element is present in all organic

Chemistry
1 answer:
faust18 [17]4 years ago
8 0
I'm pretty sure hydrogen is but i would go with #3 carbon
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A 0.010 M aqueous solution of a weak acid HA has a pH of 4.0. What is the degree of ionization of HA in the solution?
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I think the answer might be D
6 0
3 years ago
Please help!!! I have to submit this in a few!!!​
Effectus [21]

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c

Explanation:

3 0
3 years ago
Which law is known as the law of action-reaction
kakasveta [241]

Answer:

Newton's third law

Explanation:

The law of action-reaction (Newton's third law) explains the nature of the forces between the two interacting objects. According to the law, the force exerted by object 1 upon object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 upon object 1.

6 0
4 years ago
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The Safe Drinking Water Act (SDWA) sets a limit for mercury-a toxin to the central nervous system-at 0.002 mg/L. Water suppliers
Tcecarenko [31]

Answer:

3.75 * 10^7 g of this water would have been consumed.

Explanation:

ppm represents parts per million.

ppm = (mass solute / mass solution ) *10^6

⇒It is a way to express the concentration of very dilute solutions.

For this situation we have:

ppm = (grams of mercury / grams of solution ) * 10^6

0.004 = (x grams of mercury / 1g of solution) 10^6

x = 0.004 /10^6

x = 4 *10^-9 g Mercury

This mass of mercury is per gram of solution. In the next step, we can calculate the amount of solution (water plus mercury) that would have to be ingested to ingest 0.150 g of mercury

0.150 g of Hg * (1g of solution/ 4 *10^-9 g Mercury)= 3.75 *10^7

3.75 * 10^7 g of this water would have been consumed.

7 0
3 years ago
Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p
UNO [17]

Answer : The equilibrium constant K_c for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of Br_2.

\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}

\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M

Now we have to calculate the dissociated concentration of Br_2.

The balanced equilibrium reaction is,

                              Br_2(g)\rightleftharpoons 2Br(aq)

Initial conc.         1.731 M      0

At eqm. conc.      (1.731-x)    (2x) M

As we are given,

The percent of dissociation of Br_2 = \alpha = 1.2 %

So, the dissociate concentration of Br_2 = C\alpha=1.731M\times \frac{1.2}{100}=0.2077M

The value of x = 0.2077 M

Now we have to calculate the concentration of Br_2\text{ and }Br at equilibrium.

Concentration of Br_2 = 1.731 - x  = 1.731 - 0.2077 = 1.5233 M

Concentration of Br = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

K_c=\frac{[Br]^2}{[Br_2]}

Now put all the values in this expression, we get :

K_c=\frac{(0.4154)^2}{1.5233}=0.1133

Therefore, the equilibrium constant K_c for the reaction is, 0.1133

7 0
3 years ago
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