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2 years ago

What are things animals might do to survive?

Chemistry

1 answer:

ira [324]2 years ago

7 0

Answer:

Ok,for some animals to survive,then ww should first look at the kind of ecosystem they ate exposed to.

So first for some animals to survive,some of them will have to migrate,and so migration in a given set up of an ecosystem,the other animals may lack food as in they depend on the migrating animals as their prey hence their food.

Secondly,hibernation some animals may hibernate due to intensive weather or climate change which is inevitable,and this might affect other animals in such a way that may lead to decrease of their counterparts hence dying due to an imbalance as the other animals may become prey to the predators available.

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2 years ago

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What is the standard unit used to measure mass

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2 years ago

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Consider four elements from Group 7A: fluorine in the second period, chlorine in the third period, bromine in the fourth period,

madam [21]

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3 years ago

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Help I'm giving you a lot of points and do not put random or crazy stuff the real question

dybincka [34]

Answer:

Endothermic

Explanation:

Because Endothermic is cold and exothermic is hot and if you are using an ice pack it would be Endothermic and if you were using something that was hot it would be exothermic

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2 years ago

15.0 moles of gas are in a 8.00 L tank at 22.3 ∘C∘C . Calculate the difference in pressure between methane and an ideal gas unde

leva [86]

Answer:

$\Delta P=4.10atm$

Explanation:

Hello!

In this case, since the ideal gas equation is used under the assumption of no interaction between molecules and perfectly sphere-shaped molecules but the van der Waals equation actually includes those effects, we can compute each pressure as shown below, considering the temperature in kelvins (22.3+273.15=295.45K):

$P^{ideal}=\frac{nRT}{V}=\frac{15.0mol*0.08206\frac{atm*L}{mol*K}*295.45K}{8.00L}=45.5atm$

Next, since the VdW equation requires the molar volume, we proceed as shown below:

$v=\frac{8.00L}{15.0mol}=0.533\frac{L}{mol}$

Now, we use its definition:

$P^{VdW}=\frac{RT}{v-b} -\frac{a}{v^2}$

Thus, by plugging in we obtain:

$P^{VdW}=\frac{0.08206\frac{atm*L}{mol*K}*295.45K}{0.533mol/L-0.0430L/mol} -\frac{2.300L^2*atm/mol^2}{(0.533L/mol)^2}\\\\P^{VdW}=49.44atm-8.09atm\\\\P^{VdW}=41.4atm$

Thus, the pressure difference is:

$\Delta P=45.5atm-41.4atm\\\\\Delta P=4.10atm$

Best regards!

6 0

2 years ago