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Shtirlitz [24]
2 years ago
9

Which of the following terms describes an accumulation of rocky, sandy, or clayey material deposited at the end of a glacier?

Chemistry
1 answer:
OLEGan [10]2 years ago
3 0

Answer:

Option (D)

Explanation:

<u>Eskers are the long ridges that are comprised of rocks, sands and clay particles and are deposited towards the end of the glaciers</u>. These are fluvioglacial depositional features. These particles are exposed after the glaciers recede. These ridges are formed parallel to the earlier flow direction of ice. The size of eskers is generally smaller as it carries smaller particles such as rocks, sands, and gravels, in comparison to the different type of moraines. It is because the flow velocity decreases as the glaciers melt. So, these eskers are formed at the end of the glaciers.

Thus, the correct answer is option (D).

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3 years ago
A gas has a volume of 1.75L at -23C and 150.0kPa. At what temperature would the gas occupy 1.30L at 210.0kPa?
Katen [24]

Answer:

T2 = 260 K  

Explanation:

<em>Given data:</em>

P1 = 150.0 k Pa

T1 = (-23+ 273.15) K = 250.15 K  

V1 = 1.75 L  

P2 = 210.0 kPa  

V2 = 1.30 L

<em>To find:</em>

T2 = ?

<em>Formula:</em>

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

T_2 = \frac{P_2 V_2 T_1}{P_1 V_1}

<em>Calculation:</em>

T2 = (210.0 kPa) x (1.30 L) x (250.15 K) / (150.0 kPa) x (1.75 L)

T2 = 260 K  

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3 years ago
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pishuonlain [190]
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7 0
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The pH of a 0.150 molar solution of a weak acid is 4.10. What is the pKa of the acid?
Kamila [148]
Answer: 4.21×10⁻⁸

Explanation:


1) Assume a general equation for the ionization of the weak acid:

Let HA be the weak acid, then the ionization equation is:

HA ⇄ H⁺ + A⁻

2) Then, the expression for the ionization constant is:

Ka = [H⁺][A⁻] / [HA]

There, [H⁺] = [A⁻], and [HA] = 0.150 M (data given)


3) So, you need to determine [H⁺] which you do from the pH.

By definition, pH = - log [H⁺]

And from the data given pH = 4.1


⇒ 4.10 = - log [H⁺] ⇒ [H⁺] = antilog (- 4.10) = 7.94×10⁻⁵

4) Now you have all the values to calculate the expression for Ka:

ka = 7.94×10⁻⁵ × 7.94×10⁻⁵ / 0.150 = 4.21×10⁻⁸
3 0
3 years ago
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