Isotonic compound have high melting point. Why ? Name the ions present in CaS.

How might an increasing population of robins positively and negatively affect a community?

Oliga [24]

All populations of living things are interrelated. When one population of animals, plants, or insects increase or decrease, other populations of living things are also affected. For example, when shrubs and brushy areas are removed from an ecosystem, the rabbit population will likely go down. The reduced rabbit population will lower predator populations that use rabbits as a food source.

In another example, let's assume all the dead hollow trees are removed from a forest ecosystem. Cavity nesting animals such as bluebirds, nuthatch, wrens, screech owls, squirrels and woodpeckers have very little, if any, shelter available. The number of animals of this type would be reduced. Insect populations could increase because of fewer insect eating birds and trees and other plants could be negatively affected. The whole ecosystem is affected.

The amount of suitable habitat for a species of wildlife will determine the number of animals that can survive in the area. Human activity has the greatest impact on the amount and quality of wildlife habitat in Illinois. Wildlife habitat can be destroyed or its quality diminished as a result of urban sprawl, agricultural practices, pollution, sedimentation, or habitat fragmentation.

People can also have a positive impact on wildlife populations through improvement and protection of habitat or ecosystems. The planting of trees and shrubs, as well as wildlife food plots, in the appropriate locations is one way landowners can improve wildlife habitat. People can protect ponds, streams, rivers and wetlands from sedimentation by reducing soil erosion on lands surrounding these aquatic ecosystems. Nesting boxes placed in ecosystems that lack dead, hollow trees will enhance the habitat for cavity nesting animals. There are many things people can do to improve habitat for wildlife.

There are a number of natural resource management agencies in Illinois that can help people enhance and protect wildlife habitat. Listed below are a few of the public agencies which can provide ecosystem management expertise to people in Illinois.

5 0

3 years ago

What would happen if the sand dunes in an area were destroyed?

Komok [63]

Answer:

<u>Our beaches would be unprotected</u>

In the short-term, these artificial sand hills will be destroyed by the elements. Because sand dunes protect inland areas from swells, tides, and winds, they must be protected and defended like national treasures. ... The ocean and the wind can have an unpredictable, destructive force on coastal regions.

- surfertoday

Natural sand dunes play a vital role in protecting our beaches, coastline and coastal developments from coastal hazards such as erosion, coastal flooding and storm damage. Sand dunes protect our shorelines from coastal erosion and provide shelter from the wind and sea spray.

- Waikato Regional Council

3 0

2 years ago

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Need help with chemistry problem

sukhopar [10]

Answer:

b option sir sidd dichhi good book goo

6 0

2 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

b)

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

Chlorine pentafluoride gas is collected at -17.0 °C in an evacuated flask with measured volume of 35.0 L. When all the gas has b

Svet_ta [14]

Answer:

1. The mass of Chlorine pentafluoride, ClF5, is 39.16g

2. The number of mole of Chlorine pentafluoride, ClF5, is 0.3mole

Explanation:

1. To solve the mass of Chlorine pentafluoride, ClF5, first, let us calculate the molar mass of ClF5. This is illustrated below:

Molar Mass of ClF5 = 35.5 + (5 x 19) = 35.5 + 95 = 130.5g/mol

From the ideal gas equation:

PV = nRT (1)

Recall:

Number of mole(n) = mass (m) /Molar Mass(M)

n = m/M

Now substituting the value of n in equation 1, we have:

PV = nRT

PV = mRT/M

Now we can obtain the mass of Chlorine pentafluoride ClF5 as follow:

Data obtained from the question include:

T (temperature) = -17.0 °C = - 17 + 273 = 256K

V (volume) = 35L

P (pressure) = 0.180 atm

R (gas constant) = 0.082atm.L/Kmol

m (mass of Chlorine pentafluoride) =?

M (molar mass of Chlorine pentafluoride) = 130.5g/mol

PV = mRT/M

0.180 x 35 = m x 0.082 x 256/ 130.5

Cross multiply to express in linear form as shown below:

m x 0.082 x 256 = 0.180x35x130.5

Divide both side by 0.082 x 256

m = (0.180x35x130.5) /(0.082x256)

m = 39.16g

Therefore, the mass of Chlorine pentafluoride, ClF5, is 39.16g

2. The number of mole of ClF5 can be obtained as follow:

Mass of ClF5 = 39.16g

Molar Mass of ClF5 = 130.5g/mol

Mole of ClF5 =?

Number of mole = Mass /Molar Mass

Mole of ClF5 = 39.16/130.5g

Mole of ClF5 = 0.3mole

8 0

3 years ago

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