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2 years ago

Seasonal Changes of Animals in Florida Seasonal Changes of Animals in Wyoming

1 answer:

3 0

what is different about these two different foxes is the fox in Florida looks like he is on a diet of plants. The fox in wyoming looks like he has a squirel or some kind of animal in his mouth. Also their environment that they live in is completely different from each other hope it helps

Explanation:

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What is the mass of 6.14x10^25 atoms of gold

Alexxx [7]

3452.6^15 atoms of gold.

3 0

3 years ago

What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

3 0

3 years ago

A dark purple liquid is diluted by adding water.

alexandr402 [8]

Answer: The process is diffusion.

Explanation:

a dark liquid will have the highest concentration of dark purple atoms/molecules so it gives a concentrated colour.

when water is added to it, the colourless water molecules fill up the gaps between the purple particles and so their colour fades and becomes lighter and lighter as we add more water. see the image attached where imagine the red particles are water and the blue particles are purple particles. thats why the colour fades.

6 0

1 year ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

4 0

3 years ago

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