How many significant figures are in the number 20300?

Clouds are formed by the ________ water

tatuchka [14]

Clouds are made of water. When water evaporates it turns into clouds. So, I think the answer is EVAPORATED water.

4 0

2 years ago

Why do you think scientists claim to offer only supporting evidence for a theory and not proof?

larisa [96]

Because they're not confident enough to confirm it's indeed true but they will support it for they believe it's correct.

8 0

2 years ago

Under standard-state conditions, which of the following half-reactions occurs at the cathode during the electrolysis of aqueous

Slav-nsk [51]

Answer:

The following reaction will occur at cathode:

$Ni^{+2}(aq)+2e--->Ni(s)$

Explanation:

The two half reaction during electrolysis of aqueous nickel sulfate will be

a) anode reaction :

Water will undergo oxidation and will evolve oxygen gas at anode as shown in the given reaction:

$2H_{2}O(l) ----> O_{2}(g) + 4H^{+}(aq) + 4e$

b) Cathode reaction: The reduction of Nickel ion will occur by gain of two electrons as shown in the given equation:

$Ni^{+2}(aq)+2e--->Ni(s)$

Thus the overall reaction will be:

$2H_{2}O(l) + 2Ni^{+2}----> O_{2}(g) + 4H^{+}(aq) + 2Ni(s)$

6 0

2 years ago

Help............???????

balandron [24]

B. .175

Hope it helps!

4 0

2 years ago

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

2 years ago

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How many significant figures are in the number 20300?​

How many significant figures are in the number 20300?