What is mass wasting and how can it disturb the ecosystem ? (Need this ASAP PLEASEE !!!)

What 2 ways that humans have contributed to desertifaction

sdas [7]

Deforestation and inappropriate agriculture otherwise known as over farming are factors of and causes of desertification.

8 0

3 years ago

A 50.0 mL sample of gas at 20.0 atm of pressure is compressed to 40.0 atm of pressure at constant temperature. What is the new v

Slav-nsk [51]

Answer:

20(50) = 40(V₂)

⇒1000 = 40V₂

⇒V₂ = 1000/40

⇒ V₂ = 25 mL

Explanation:

6 0

2 years ago

Answer the following questions:

Vera_Pavlovna [14]

Answer:

fdfdgiukjgrguhjnfvthhhdqddwsvhjjj

4 0

3 years ago

Read 2 more answers

PLEASE ANSWER BOTH QUESTIONS I WILL MARK YOU BRAINLIEST IF IT IS CORRECT!!

seropon [69]

Answer:

1st Question: A

2nd Question: B

Explanation:

The 1st answer would be A because if a sample is at absolute zero then the sample is at its lowest temperature none of the molecules would be able to move, this is because lower temperature= lower kinetic energy.

The 2nd answer would be B because if a sample has more temperature it speeds up it has more temperature and more kinetic energy, meaning it would move faster because there is more temperature.

7 0

3 years ago

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

1. Chemical reaction:

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

2. Initial concentrations:

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

4. Equilibrium expression

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

5. Solve:

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0

3 years ago