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choli [55]
3 years ago
6

Which of the following is a way that some animals bodies adapt to living in extremely cold climates

Chemistry
2 answers:
HACTEHA [7]3 years ago
7 0
B hibernation but it deppends on the animal
ratelena [41]3 years ago
6 0

Answer : The correct answer is B ) Hibernation .

Adaptation is a process where an animal well suited in surrounding area with time . Animals living in extremely cold habitat adapts their body which can protect them against cold . If the options are checked :

A) Clusters : Clusters is adaptation done by marine animals . They form clusters which are called as " Beds " on surfaces which provide shelter to many plants .

B) Hibernation : It is state of inactivity or winter dormancy . This is extended sleep by animals by reducing their metabolic rates and lowering of body temperature in scarcity of food in extreme cold winters. SO we can Hibernation is adapted by cold animals to survive from extreme cold.

C) Burrows : Burrows are tunnels dug by animals basically by desert animals to protect themselves from scorching heat or by animals for protection .

D) Estivation : This is similar to hibernation , state of dormancy but in very hot and dry season . Animals during summer ( hot and dry ) go to underground where the environment is cooler to protect themselves from excessive loss of water . So it is adaptation of desert animals ,

Hence Hibernation is correct adaptation for cold animals .

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The answer is the most probable location of electrons in an atom
8 0
2 years ago
Calculate the mass (in grams) of 0.473 mol of titanium
xz_007 [3.2K]

Explanation:

n=given mass ÷molar mass

make given mass become the subject of the formula by

multiplying the molar mass on both sides of the equation.

n=0.473mol

given mass=??

molar mass=48

therefore,given mass=n×molar mass

=0.473×48

=22.704grams

mass in grams is 22.704grams

7 0
3 years ago
An increase in the surface area of reactants in a heterogeneous reaction will result in what?
marusya05 [52]
It will result in an increase in the rate of rxn
4 0
2 years ago
Calculate the molecular weight of a substance. In which the solution of this substance in the water has a concentration of 7 per
Eduardwww [97]

Answer : The molecular weight of a substance is 157.3 g/mol

Explanation :

As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.

Mass of solute = 7 g

Mass of solution = 100 g

Mass of solvent = 100 - 7 = 93 g

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=k_f\times\frac{\text{Mass of substance(solute)}\times 1000}{\text{Molar mass of substance(solute)}\times \text{Mass of water(solvent)}}

where,

\Delta T_f = change in freezing point

T_f^o = temperature of pure water = 0^oC

T_f = temperature of solution = -0.89^oC

K_f = freezing point constant of water = 1.86^oC/m

m = molality

Now put all the given values in this formula, we get

(0-(-0.89))^oC=1.86^oC/m\times \frac{7g\times 1000}{\text{Molar mass of substance(solute)}\times 93g}

\text{Molar mass of substance(solute)}=157.3g/mol

Therefore, the molecular weight of a substance is 157.3 g/mol

7 0
3 years ago
Predict the missing product of this equation<br><br><br>1 MgF2 + 1 Li2CO3 -&gt; 1 ______ +2LiF
ch4aika [34]

Answer:

MgCO₃

Explanation:

From the question given above, we obtained:

MgF₂ + Li₂CO₃ —> __ + 2LiF

The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:

MgF₂ (aq) —> Mg²⁺ + 2F¯

Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯

MgF₂ + Li₂CO₃ —>

Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:

MgF₂ + Li₂CO₃ —> __ + 2LiF

MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF

Therefore, the missing part of the equation is MgCO₃

8 0
2 years ago
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