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2 years ago

Write a hypothesis about the effect of dry conditions on earthworm behavior. Use the "if . . . then . . .

Chemistry

2 answers:

alexgriva [62]2 years ago

5 0

Answer:

If the earthworm's skin dries then they will not breath properly because moisture help them out to ease the diffusion of oxygen through its skin.

Explanation:

Hello,

Based on the required hypothesis, the best describing one considering the question "How is earthworm behavior affected by external stimuli?", would be:

If the earthworm's skin dries then they will not breath properly because moisture help them out to ease the diffusion of oxygen through its skin.

In this case, we are specifying a possible condition, this is "the earthworm's skin dries" which is a cause of an external stimuli (dryness). Next, we propose an effect, this is "they will not breath properly", that is caused by the initial part. Finally, we explain the effect via the cause "moisture help them out to ease the diffusion of oxygen through its skin" because is something testable via the subsequent experiments when considering the scientific method.

Best regards.

Vlada [557]2 years ago

4 0

sorry about the late response...

<u>If an earthworm is exposed to dry conditions, then it will retreat to a moist place because its skin needs to stay moist for the earthworm to survive.</u>

Stasia

2 years ago

who u

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Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

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