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rodikova [14]
3 years ago
10

Predict what might happen if the human body did not have specialized cells, tissues, organs, and organ systems to maintain homeo

stasis
Chemistry
2 answers:
sukhopar [10]3 years ago
6 0
<span>The body would be unable to maintain homeostasis because the cells of the body would have too many jobs to do. The cells likely would not be able to do all of these jobs.
</span><span>It wouldn't be able to live because the organ system helps you breath.
</span>The specialisation of cells is paramount to having a functional multicellular <span>organism.
If the body didn't specialise then it wouldn't be an organism; it </span>would just be a colony of cells much like those that bacteria and other single <span>celled organisms form.
</span>

Hope this helps.
bekas [8.4K]3 years ago
3 0
We would be dead. No heart, brain or blood flow <span />
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Answer:

2

Explanation:

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What is the chemical formula for ammonium oxalate
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A process at constant T and P can be described as spontaneous if ΔG &lt; 0 and nonspontaneous if ΔG &gt; 0. Over what range of t
creativ13 [48]

Answer:

Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

                              2 SO3(g)   →          2 SO2(g) + O2(g)

ΔHf° (kJ mol-1)  -395.7                        -296.8

S° (J K-1 mol-1)  256.8                         248.2              205.1

ΔH° =  kJ

S° =  J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine  ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where  ν  is the stoichiometric coefficient in the balanced chemical equation.

For ΔS we have likewise

ΔrxnS =  ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) =  2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ =  197.8 - T(0.1879)

and the reaction will become spontaneous when the term  T(0.1879)  becomes greater that 197.8,

0 = 197.8 - 0.1879 T  ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

4 0
3 years ago
When a connector is marked with "al-cu," the connector is suitable for use with copper, copper-clad aluminum, and aluminum condu
IgorLugansk [536]

When connectors are marked with a combination of metals, it can be used as a connector of one of the metals or an alloy of the two metals. So in this case, since the marking is “Al – Cu” where Al is aluminium and Cu is copper, therefore the answer is:

<span>Yes, it is suitable for use with copper, copper-clad aluminum, and aluminum conductors.</span>

6 0
3 years ago
Please help with this chemistry question it’s due by midnight
Scilla [17]

Answer:

HF is the limiting reactant

Explanation:

The balanced equation for the reaction is given below:

SiO₂ + 4HF —> SiF₄ + 2H₂O

From the balanced equation above,

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Finally, we shall determine the limiting reactant. This can be obtained as illustrated below:

From the balanced equation above,

1 mole of SiO₂ reacted with 4 moles of HF.

Therefore, 7.5 moles of SiO₂ will react with = 7.5 × 4 = 30 moles of HF.

From the calculation made above, we can see clearly that it will take a higher amount (i.e 30 moles) of HF than what was given from the question (i.e 5 moles) to react completely with 7.5 moles of SiO₂.

Therefore, HF is the limiting reactant and SiO₂ is the excess reactant.

3 0
3 years ago
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