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3 years ago

15

One of the human body's adaptation is to sweat when it gets too hot. Sweat glands in the glands in the skin produce a solution o

f mostly water that __________ through pores at the skin's surface. A. spurts B. shivers C. boils D. evaporates

1 answer:

Rina8888 [55]3 years ago

5 0

The answer is D. Evaporated

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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lo

riadik2000 [5.3K]

Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

$(1):C_6H_6(g)(425.0K)\rightarrow C_6H_6(g)(353.0K)\\\\(2):C_6H_6(g)(353.0K)\rightarrow C_6H_6(l)(353.0K)\\\\(3):C_6H_6(l)(353.0K)\rightarrow C_6H_6(l)(335.0K)$

The expression used will be:

$\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat released by the reaction = ?

m = mass of benzene = 125 g

$c_{p,g}$ = specific heat of gaseous benzene = $1.06J/g^oC$

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

$\Delta H_{vap}$ = enthalpy change for vaporization = $33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g$

Molar mass of benzene = 78.11 g/mole

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

$\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]$

$\Delta H=-67682.5J=-67.7kJ$

Therefore, the energy removed must be, -67.7 kJ

4 0

3 years ago

The reaction for producing glucose in plants, called photosynthesis, is:

Firlakuza [10]

Answer:

41.16 moles of H2O

Explanation:

Ratio for the products-reactants is 1:6, so 1 mol of glucose is produced when plants use 6 moles of water.

Then, let's make a rule of three:

1 mol of glucose is produce by using 6 moles of water

6.86 moles of glucose are produced by the use of (6 . 6.86)/1 = 41.16 moles of H2O

8 0

3 years ago

Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl

tigry1 [53]

Answer:

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

<em>∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) </em>= (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

Calculating ∆S°rxn:

∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants

<em>∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) </em>= (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) =<em> - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.</em>

<em></em>

Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

<em>∴ ∆G°rxn = ∆H°rxn - T∆S°rxn </em>= (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = <em>- 10.555 kJ/mol.</em>

4 0

3 years ago

You have 16.0 g of some compound and you perform an experiment to remove all of the oxygen, 11.2 g of iron is left. What is the

Makovka662 [10]

The empirical formula of this compound is $Fe_2O_3$

<h3>Empirical formula </h3>

To calculate the empirical formula of a compound, the value of moles of each element is needed.

As we have the information of the mass value, we will use the molar mass expression, which corresponds to:

$MM_O = 16g/mol\\MM_Fe = 55.8g/mol$

$MM = \frac{m}{mol}$

O

$16 = \frac{4.8}{x}$

$x = 0.3mol$

Fe

$55.8=\frac{11.2}{x}\\x = 0.2$

As the value of the empirical formula must be an integer, simply multiply the two values by a common factor:

$O = 0.3 \times 10 = 3\\Fe = 0.2 \times 10 = 2$

$Fe_2O_3$

So, the empirical formula of this compound is $Fe_2O_3$ .

Learn more about empirical formula: brainly.com/question/1247523

3 0

2 years ago

Is Bromothymol Blue acidic or basic?

notka56 [123]

It is mostly used in applications that need measuring substances that would have a<span> relatively neutral pH . </span>A<span> common use is for measuring the presence of carbonic </span>acid<span> in </span>a<span> liquid. so yes its acidic</span>

8 0

3 years ago