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DanielleElmas [232]
3 years ago
11

7. A species is _ when its population has become so low that it is close to becoming extinct

Chemistry
1 answer:
Anna35 [415]3 years ago
7 0

Answer:

Endangered. The meaning of endangered is when a species declines in number- this might be because they are being eaten by another animal which has a higher popupayion, or perhaps because they are being hunted by humans for their meat or fur. Neither reason might be that their food or habitat has been destroyed b humans to create land for farming or housing. one example of name endangered species is Polar Bears: they like in the Antarctic where there is no land, simply ice. Due to humans causing climate change, the world is warming up and the ice is melting. This means polar bears are dying out because it would be too tiring to swim to find ice. For the same reason, other animals are dying in the Antarctic meaning that Poplar Bears have less to eat (being at the top of the food chain -no animal eats polar bears).

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andreev551 [17]
The answer is all of the above
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Which factors affect a river's load?
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To answer the question that is: "which factors affect a river's load", we have to understand that all the things mentioned are important. All these options (river's slope, streambed shape and volume of flow) affect the amount of energy that the river has to spend and the way the river spend that energy, so, it is right to mark the alternative <span>d) all of the above.</span>
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There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc
zhuklara [117]

Answer:

K_{sp} of Zn(CH_{3}COO)_{2} is 0.0513

Explanation:

Solubility equilibrium of Zn(CH_{3}COO)_{2}:

Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}

Solubility product of Zn(CH_{3}COO)_{2} (K_{sp}) is written as-            K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}

Where [Zn^{2+}] and [CH_{3}COO^{-}] represents equilibrium concentration (in molarity) of Zn^{2+} and CH_{3}COO^{-} respectively.

Molar mass of Zn(CH_{3}COO)_{2} = 183.48 g/mol

So, solubility of Zn(CH_{3}COO)_{2} = \frac{43.0}{183.48}M = 0.234M

1 mol of Zn(CH_{3}COO)_{2} gives 1 mol of Zn^{2+} and 2 moles of CH_{3}COO^{-} upon dissociation.

so,   [Zn^{2+}] = 0.234 M and [CH_{3}COO^{-}] = (2\times 0.234)M=0.468M

so, K_{sp}=(0.234)\times (0.468)^{2}=0.0513          

8 0
3 years ago
Your blood contains many dissolved solids. What do you think could be done if you needed to remove the water from a sample of bl
s2008m [1.1K]

Answer:

could the answer be boil the water away?

Explanation:

if the water gets boiled and evaporates, than you are left with the solids

8 0
3 years ago
A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t
andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)

<u>Where:</u>

<em>μ (l): is the chemical potential of 2-propanol in solution    </em>

<em>μ° (l): is the chemical potential of pure 2-propanol   </em>

<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>

<em>T: is the temperature = 82.3 °C = 355.3 K </em>

<em>x: is the mole fraction of 2-propanol = 0.41 </em>

\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)

\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}

\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}  

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.    

I hope it helps you!    

8 0
3 years ago
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