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2 years ago

What is the difference between Temperate grassland and Desert????????

Chemistry

2 answers:

Alex2 years ago

8 0

Answer:

i think one has grass

Explanation:

arsen [322]2 years ago

4 0

what is the difference between Temperate grassland and Desert??

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The ksp value for lead(ii chloride is 2.4 × 10?4. what is the molar solubility of lead(ii chloride?

charle [14.2K]

Molar solubility is number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.

The molar solubility of lead(ii) chloride with ksp value of 2.4 × 10e4 can be solve as:

Ksp = s2 = 2.4 × 10e4
s2 = 2.4 × 10e4
s = √(2.4 × 10e4)
s = 154.9 mol/L

6 0

3 years ago

In the heating curve of an unknown substance, the line labeled C represents what change(s)?

IceJOKER [234]

Answer:

C) Temperature and Kinetic Energy.

Explanation:

Hello there!

In this case, according to the generic heating curve on the attached file, it possible to see that on the point C, whereas the line is diagonal, the temperature increases, but also the kinetic energy increases because the molecules gain energy due to the increase of the temperature. It is important to say that on flat lines, like those on B and D, the phase change takes place and just the potential energy change.

In such a way, we infer that the answer is C) Temperature and Kinetic Energy.

Best regards!

6 0

3 years ago

Read 2 more answers

Refer to the example about diatomic gases A and B in the text to do problems 19 - 27.

soldier1979 [14.2K]

Answer:a

Explanation:

5 0

3 years ago

An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid,

natima [27]

Answer:

The answer is "Option B".

Explanation:

$\to CH_3COOH + NaOH \longleftrightarrow CH_3COONa + H_2O\\\\\to CH_3COONa + NaOH\longleftrightarrow CH3COONa\\\\\therefore \ mol\ NaOH = (5 \ E-3\ L)\times(0.10 \ \frac{mol}{L}) = 5 \ E-4\ mol\\\\$

$\to mol\ CH_3COOH = (0.05 \ L)\times(0.20 \frac{mol}{L}) = 0.01 \ mol\\\\\to C \ CH_3COOH = \frac{(0.01 \ mol - 5 \ E-4\ mol) }{(0.105 \ L)}\\\\\to C \ CH_3COOH = 0.0905 \ M\\\\\therefore \ mol \ CH_3COONa = (0.05\ L )\times (0.20 \ \frac{mol}{L}) = 0.01 \ mol\\\\$

$\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\$

$\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7$

5 0

3 years ago

Question 10 (1 point)

Strike441 [17]

A. Fluorine typically gains 1 electron, like the rest of the halogens.
b. The charge is -1, since it gains 1 electron.
c. Anion, since the charge is negative.

8 0

3 years ago