All categories

2 years ago

Do you think precipitation has a greater impact on the rate of chemical weathering or mechanical weathering? Explain

Chemistry

1 answer:

nexus9112 [7]2 years ago

6 0

Greater rainfall has an greater increase on the rate of chemical weathering. Rain is a form of precipitation.

You might be interested in

What is liquor ammonia fortis?

erik [133]

Answer:

Ammonia fortis liquor is a saturated solution of ammonia in water. It is also called 880 ammonia. Its relative density is 0.880. It is stored in tightly sealed bottles in a cold place. (Sorry if I'm wrong)

Explanation:

6 0

3 years ago

Read 2 more answers

Please help me:( I’ll give u points of its right

goldenfox [79]

Answer:

I believe.. ............

8 0

3 years ago

If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? A. one-fourth B.

Basile [38]

$m=m_{0}*(\frac{1}{2})^{\frac{60}{30}}\\\\
m=m_{0}*(\frac{1}{2})^{2}\\\\
m=\frac{1}{4}m_{0}$

If the half-life of a sample of a radioactive substance is 30 seconds, how much would be left after 60 seconds? <span>A. one-fourth</span>

7 0

3 years ago

Read 2 more answers

Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a

svetoff [14.1K]

Answer:

The answer to your question is P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

= 373°K

Formula

$(P + \frac{a}{v^{2}} )(v - b) = RT$

Substitution

$(P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)$

Simplify

(P + 0.0094)(22.3829) = 30.586

Solve for P

P + 0.0094 = $\frac{30.586}{22.3829}$

P + 0.0094 = 1.366

P = 1.336 - 0.0094

P = 1.357 atm

7 0

4 years ago

What volume (in liters) does 3.91 moles of nitrogen gas at 5.35 atm and 323 K occupy

daser333 [38]

Moles = n = 3.91 mol

Pressure = P = 5.35 atm

Temperature = T = 323 K

Volume = V = ?

Formula used: Ideal Gas Equation is used,

P V = n R T
Solving for V,
V = n R T / P
Putting Values,
V = (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm

V = 19.36 L

5 0

3 years ago