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3 years ago

A strike slip fault is a result of

Chemistry

1 answer:

11111nata11111 [884]3 years ago

8 0

<span>I think its horizontal compression!

Plz give me Brainlyest if i got it right?

</span>

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If a gaseous mixture is made by combining 2.74 g Ar 2.74 g Ar and 4.90 g Kr 4.90 g Kr in an evacuated 2.50 L container at 25.0 ∘

krok68 [10]

Answer:

Hfhfj

Explanation:

7 0

3 years ago

calculate the ph of a 0.020 m carbonic acid solution, h2co3(aq), that has the stepwise dissociation constants ka1

fgiga [73]

The calculated pH is 3.79. therefore, the solution is acidic.

No, carbonic acid is not a strong acid. H2CO3 is a weak acid that dissociates into a proton (H+ cation) and a bicarbonate ion (HCO3- anion). This compound only partly dissociates in aqueous solutions.

H2 CO3 = H (+) + HCO3(-) Ka1 = 4.3 * 10^ -7

0.06 - x x x

Ka1 = x^2 / (0.06 - x) = 4.3 * 10^ - 7

A low Ka => x << 0.06 => 0.06 -x ≈ 0.06

=> Ka1 ≈ x^2 / 0.06 => x^2 ≈ 0.06 * Ka1 = 0.06 * 4.3 * 10^-7

=> x ≈ √ [ 2.58 * 10 ^ -8] = 1.606 * 10^ - 4 = 0.0001606

Second dissociation

HCO3(-) = H (+) + CO3(2-) Ka2 = 5.6 * 10^ - 11

0.0001606 - y y y

Ka2 ≈ y^2 / 0.0001606 => y = √ [0.0001606 * 5.6* 10^ -11]

y = 9.48 * 10^ -8

An acidic solution has a high concentration of hydrogen ions (H +start superscript, plus, end superscript), greater than that of pure water.

[H+] = x + y = 1.607 * 10^ -4

pH = - log [H+] = 3.79

Learn more about concentration here-

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5 0

2 years ago

Draw a structure for the product of nucleophilic substitution obtained on solvolysis of tert−butyl bromide in methanol, and arra

Elena-2011 [213]

Answer:

2-methoxy-2-methylpropane

Explanation:

The first step for this reaction is the carbocation formation. In this step, a tertiary carbocation is formed. Also, we will have a good leaving group so bromide will be formed. Then the methanol acts as a nucleophile and attacks the carbocation. Next, a positive charge is generated upon the oxygen, this charge can be removed when the hydrogen leaves the molecule as $H^+$ . (See figure)

6 0

4 years ago

Suppose a 20.0 g gold bar at 35.0°C absorbs 70.0 calories of heat energy. Given that the specific heat of gold is 0.0310 cal/g °

timofeeve [1]

We know, change in temperature is given by :

$T_2-T_1=\dfrac{q}{mC_p(Gold)}$

Putting all given values, we get :

$T_2-T_1=\dfrac{70\ cal}{20\ g\times 0.0310\ cal/g^o\ C}\\\\T_2-T_1=112.90^oC\\\\T_2-35^oC=112.90^oC\\\\T_2=(112.90+35)^oC\\\\T_2=147.9^oC$

Hence, this is the required solution.

6 0

4 years ago

A student forgot to read the label on the jar carefully and put potassium chloride in the crucible instead of potassium chlorate

lys-0071 [83]

<span>The object that was trying to be oxidized would end up being reduced. There would be no net reaction otherwise. The KCl would have simply melted after a long enough time and with the application of enough heat to the crucible.</span>

7 0

3 years ago

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