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DerKrebs [107]
4 years ago
10

What is the net charge of arginine in a solution of ph 1.0?

Chemistry
1 answer:
taurus [48]4 years ago
3 0

Answer: +2

Explanation:

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According to the rock cycle, which of the following transitions are possible? Assume an unlimited number of steps.
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(D)All of the above are possible.

Explanation:

According to the rock cycle, all of the transitions given are possible with several unlimited steps.

The rock cycle shows how the different rock types are transformed from one form to another.

In the rock cycle the major rock types are taken into consideration. These rock types are Igneous, sedimentary and metamorphic.

  • An intrusive igneous rock becomes a sedimentary rock.

Based on the rock cycle, an intrusive rock can become a sedimentary rock. This is igneous to sedimentary rock conversion. Here, the intrusive rock is brought during a terrain change to the surface. Weathering, erosion, transportation and deposition combines to produce sediments and take them to their basin of deposition where they form sedimentary rocks.

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  • A metamorphic rock becomes an extrusive igneous rock.

When metamorphic rocks are subjected to intense heat and pressure, they reach their melting point.

As they melt, they produce magma which can be brought to the surface to form extrusive volcanic rocks.

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6 0
3 years ago
Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with
stepan [7]

Answer:

1) Net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2) 0.765 M is  the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq) ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq) ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq) ..[3]

H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)

From one:[1] ,[2] and [3]:

2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)

Cancelling common ions on both sides to get net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2CO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL

Putting values in above equation, we get:

M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M

0.765 M is  the molarity of the carbonic acid solution.

6 0
3 years ago
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