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3 years ago

Describe cold blooded animal.also some five example.

Chemistry

2 answers:

Vlad [161]3 years ago

3 0

Answer:

Cold-blood animals are the animals that are not capable of regulating their body temperature,Cold blood animals included reptiles,fishes,amphibians,insects and other invertebrates.

Ilya [14]3 years ago

3 0

Answer:

Cold blooded animals are animals which stay in water

Explanation:

Examples are;snakes,tortoises,turtles,crocodiles,alligators

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A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol

katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of $0.80^oC$ and a freezing point depression constant $K_f=7.82^oC.kg/mol$ . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

Answer: The freezing point of the solution is $-17.6^oC$

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $0.80^oC$

Freezing point of solution = $?^oC$

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

$K_f$ = freezing point depression constant = $7.82^oC/m$

$m_{solute}$ = Given mass of solute (iron (III) chloride) = 81.1 g

$M_{solute}$ = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

$w_{solvent}$ = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

$0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC$

Hence, the freezing point of the solution is $-17.6^oC$

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3 years ago

Describe cold blooded animal.also some five example.​

Describe cold blooded animal.also some five example.