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3 years ago

What has the biggest impact on a region,weather or atmosphere.

Chemistry

1 answer:

andreev551 [17]3 years ago

5 0

Weather.

You see how the weather is in a certain place can determine how the plant life and just life in general would live. It can also effect people. Let's say there was a drought somewhere, this would effect people whereas, they could lose a significant source of water and food because it would make them likely have to find a new water source or the animals would. Weather also effects the land in harmful ways. For example when it floods, the ground gets muddy but can also get so muddy below a certain area, that the soil entirely collapses into the ground (that would be a sinkhole).

There are picture below to show you some of the effects.

-R3TR0 Z3R0

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40 points pls (10 more if its not plaigerized plsss) thank youuuu (if u just do it for points u will get reported bc its not the

mars1129 [50]

Answer:

The freshwater sources that are generally in continuous motion and follow a defined path are called streams and rivers.

If I were to improve the lab then I will make the following changes:

The experiment aimed to observe and model the effects of rivers on erosion. So, I can make a virtual model of the river and can compare the velocity, gradients and volume of rivers.

Comparison between the low and high factors listed can help in computing the effect of the powerful river on erosion.

The high velocity. gradient and volume of the river will cause more erosion as it exerts more force.

The low volume, gradient and velocity river will affect in a less manner on erosion.

Explanation:

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5 0

2 years ago

When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 9.925 g. If the density of

Vlada [557]

Measured volume = 10 ml
mass = 9.925 g
density = 0.9975 g/ml
density = $\frac{mass}{volume}$
so actual volume = $\frac{mass}{density}$ = $\frac{9.925}{0.9975}$ = 9.95 mL
Percentage Error = $\frac{measured volume - Actual volume}{actual volume}$ x 100
= $\frac{10 - 9.95}{9.95}$ x 100 = 0.5 % error

5 0

3 years ago

What characteristics are used to clarify an area as a wetland?

Nadya [2.5K]

Soil covered, saturated, submerged, flooded w water, standing water

5 0

3 years ago

Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc

svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ, ΔSºrxn = 0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG = 1.8 x 10⁴ J

K = 8.2 x 10⁻²

Explanation:

a) C6H5−CH2CH3 ⇒ C6H5−CH=CH2 + H₂

ΔHf kJ/mol -12.5 103.8 0

ΔGºf kJ/K 119.7 202.5 0

Sº J/mol 255 238 130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate ΔHºrxn we have

ΔHºrxn = ΔHfº C6H5−CH=CH2 + ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn = 103.8 kJ + 0 kJ - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 + ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn= 202.5 kJ + 0 kJ - 119.7 kJ = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn = ΔHrxn -TΔS

we see that will happen when the term TΔS becomes greater than ΔHrxn since ΔS is positive , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that ΔºHrxn and ΔSºrxn remain constant at the higher temperature and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K = (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

ΔGrxn = ΔHrxn -TΔS

to calculate ΔGrxn with the assumption that ΔHº and ΔSºremain constant.

ΔG = 116.3 kJ - (600+273 K) x 0.113 kJ/K = 116.3 kJ - 873 K x 0.113 kJ/K

ΔG = 116.3 kJ - 98.6 kJ = 17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K x 873 K)) = 8.2 x 10⁻²

8 0

3 years ago

A gas has a density of 1.57 g/L at 40.0 Â°C and 2.00 atm of pressure. What is the identity of the gas?

Naddika [18.5K]

Answer:

Neon

Explanation:

Step 1: Given and required data

Density of the gas (ρ): 1.57 g/L

Temperature (T): 40.0°C

Pressure (P): 2.00 atm

Ideal gas constant (R): 0.08206 atm.L/mol.K

Step 2: Convert T to Kelvin

We will use the following expression.

K = °C + 273.15 = 40.0 + 273.15 = 313.2 K

Step 3: Calculate the molar mass of the gas (M)

For an ideal gas, we will use the following expression.

ρ = P × M/R × T

M = ρ × R × T/P

M = 1.57 g/L × 0.08206 atm.L/mol.K × 313.2 K/2.00 atm

M = 20.17 g/mol

The gas with a molar mass of 20.17 g/mol is Neon.

6 0

2 years ago