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3 years ago

What are three components that make a complete description of a location?

Chemistry

1 answer:

vodka [1.7K]3 years ago

4 0

Answer:

Distance, direction and symbol.

Explanation:

Distance, direction and symbol are the three components of the map which make a complete description of a location. On the map, these three components are present which provides information about a specific location on the map. Map is a drawing on the paper that shows the geography of the whole world and provides information in detail.

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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.

6 0

3 years ago

Read 2 more answers

A sample of seawater contains 1.3g of calcium ions in 3,100kg of solution. what is the calcium ion concentration of this solutio

MakcuM [25]

Unit ppm stands for parts per million. in terms of mass, ppm is equivalent to mg/kg.
since 1 kg is 10⁻⁶ mg, 1 kg is equivalent to million mg.
therefore mg/kg is also ppm.
there are 1.3 g of Ca ions in 3100 kg
if 3100 kg contains - 1.3 g of Ca
then 1 kg contains - 1.3 g / 3100 kg
then Ca ions - 0.42 x 10⁻³ g/kg
Ca ion concentration - 0.42 mg/kg
therefore Ca ion concentration is 0.42 ppm

8 0

4 years ago

What is the specific heat of a 42.6 gram compound that changes from 12 Celsius to 46 Celsius when it absorbs 1260 KJ of energy

7nadin3 [17]

Answer:

869.93J/g°C

Explanation:

From the question given, we obtained the following:

M = 42.6g

T1 = 12°C

T2 = 46°C

ΔT = T2 — T1 = 46 — 12 = 34°C

C =?

Q = 1260 KJ = 1260000J

Q = MCΔT

C = Q / MΔT

C = 1260000 /(42.6 x 34)

C = 869.93J/g°C

Therefore, the specific heat capacity is 869.93J/g°C

5 0

4 years ago

PLS HELP THIS IS TO HARD PLS

lions [1.4K]

The correct answer is D because they are all in the same group and all elements in the same groups as each other are always very similar.

7 0

3 years ago

Which of the following substances is not a fluid?<br> air<br> iron<br> water<br> helium gas

Ann [662]

The only following substance that is not fluid is air

3 0

4 years ago

Read 2 more answers