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3 years ago

How often do most coastlines on Earth experience two high tides and two low tides?

Chemistry

1 answer:

Elza [17]3 years ago

4 0

Answer:

24 hours and 50 minutes

Explanation:

Because the Earth rotates through two tidal “bulges” every lunar day, coastal areas experience two high and two low tides every 24 hours and 50 minutes. High tides occur 12 hours and 25 minutes apart. It takes six hours and 12.5 minutes for the water at the shore to go from high to low, or from low to high.

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What kind of glacier has pieces that breaks off as icebergs

forsale [732]

There is no specific name for a glacier that break off as an iceberg. However, the part of the glacier in which this happens is called the "zone of wastage". Chunks break off in a process called "calving".

5 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

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

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

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

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

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

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

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

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

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 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

Jonathan claims that since only humans can read, it must be an inherited trait, passed from human parents to their children. Is

Crank

Jonathan is not correct because it has to be a trait that is learned.

So in that case Jonathan has to say that this is not an inherited trait it is learned by most people.

5 0

3 years ago

How many molecules (not moles of nh3 are produced from 5.01×10?4 g of h2?

Irina-Kira [14]

2,02g ------- 6,02×10²³
5,01×10⁴g --- x

$x=\frac{5,01*10^{4}g*6,02*10^{23}}{2,02g}=14,93*10^{27}$

N₂ + 3H₂ ⇒ 2NH₃
1mol : 3mol : 2mol
18,06×10²³ : 12,04×10²³
14,93×10²⁷ : y

$y=\frac{14,93*10^{27}*12,04*10^{23}}{18,06*10^{23}}\approx9,95*10^{27}$

5 0

3 years ago

How does the volume occupied by a cubic centimeter compare with the volume occupied by a millimeter

Bad White [126]

I'm pretty sure it's the same 2mm = 2cc = 2c^2

7 0

3 years ago