Name 3 variables that afect a living systom

5. All of the following statement describes compounds, EXCEPT?

Sveta_85 [38]

Answer:

C

Explanation:

Compounds cannot be separated by any physical means.

3 0

3 years ago

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In the simulation, open the Micro mode, then select the solutions indicated below from the dropdown list above the beaker. The b

SCORPION-xisa [38]

Answer:

In order of basicity we have

1. Soda pop (Least basic Normally called acidic)

2. Orange juice

3. Milk

4. Blood (slightly basic)

5. Hand soap

6. Drain cleaner (Highly basic)

Explanation:

Orange juice; the pH of orange juice is in the 3.3 to 4.2 range

Milk; the pH of milk about 6.5 to 6.7

Blood; the blood pH is around 7.35 to 7.45

Hand soap with contents such as ammonium hydroxide is basic, its pH is about 9-10

Drain cleaner contains baking soda or sodium bicarbonate which basic with a pH of 12 to 14

Soda pop pH of soda pop is in the range of 2.34 to 3.10. It contains carbonated water with a pH of 3–4, making it mildly acidic.

Arranging the above listed in order of increasing basicity, we have

1. Soda pop

2. Orange juice

3. Milk

4. Blood

5. Hand soap

6. Drain cleaner

6 0

3 years ago

Seedless plants can reproduce through asexual reproduction. T or F

Neko [114]

True..........................................................

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3 years ago

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8.03 Solutions Lab Report<br> Does anyone have a PDF or Document of FLVS 8.03 Solutions Lab

GarryVolchara [31]

8.03 solutions report is described below.

Explanation:

8.03 Solutions Lab Report

In this laboratory activity, you will investigate how temperature, agitation, particle size, and dilution affect the taste of a drink. Fill in each section of this lab report and submit it and your pre-lab answers to your instructor for grading.

Pre-lab Questions:

In this lab, you will make fruit drinks with powdered drink mix. Complete the pre-lab questions to get the values you need for your drink solutions.

Calculate the molar mass of powered fruit drink mix, made from sucrose (C12H22O11).

Using stoichiometry, determine the mass of powdered drink mix needed to make a 1.0 M solution of 100 mL.

7 0

4 years ago

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Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

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3 years ago

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