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3 years ago

What are 5 facts about the enviorment in the Atacama Desert?

2 answers:

8 0

The Atacama Desert is a plateau in South America.

Covering a 1,000-kilometre strip of land on the Pacific coast.

West of the Andes mountains. It is the driest non-polar desert in the world.

In the late 1800s, Chile and Bolivia disputed this land in the Guerra del Pacífico (War of the Pacific) becaue both countries claimed to be rightful owners of this region that.

During the day, temperatures in the desert can reach around 40º C (104º F).

boyakko [2]3 years ago

7 0

5 facts

- The Atacama Desert is located in South America
- Covers an area of approximately 128,000 square kilometers
- The Atacama Desert is home to many species of animals including insects and arachnids, reptiles and amphibians, mammals and birds.
- Many species of plant life can be found growing in the Atacama Desert including trees, shrubs, cacti, herbs and grasses.
- It is among the most driest places on Earth as it receives little to no rainfall.

Hopefully that works! :))

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Round off the following measurement to three significant figures. 1.296 g

REY [17]

It would be 1 or 2 because if the number is higher than 5 you need to round up , if its lower than 5 you need to round down.

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3 years ago

What is the conversion factor for in^3 to cm^3

DochEvi [55]

You multiply by 16.3871

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2 years ago

You are provided with a stock solution with a concentration of 1.0x10-5 M. You will be using this to make two standard solutions

artcher [175]

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

2 refers to the final state

1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.

C₁ . V₁ = C₂ . V₂

(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL

V₁ = 2.0 mL

2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.

C₁ . V₁ = C₂ . V₂

(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL

V₁ = 2.5 mL

8 0

3 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

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3 years ago

State saytzeff's rule

LUCKY_DIMON [66]

Answer:

The more substituted alkene will be the major product.

Explanation:

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3 years ago

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