State the hypothesis of continental drift.

Elis [28]

The answer is E and F.

Hope this helps!!

8 0

2 years ago

A human red blood cell is about 0.000008 meters in diameter which of the following represents this number in scientific notation

lukranit [14]

Given:

Diameter of a red blood cell = 0.000008

To determine:

The scientific notation corresponding to the given number

Explanation:

Scientific notation is a concise way of representing a very large or small number. It is written in two parts:

Given Number = Digits * Power of 10

In this case we have:

0.000008 = 8.0 * 10⁻⁶

3 0

3 years ago

A sample of water vapor has a volume of 3.15 L, a pressure of 2.40 atm, and a temperature of 325 K. What is the new temperature,

lara31 [8.8K]

Answer:

The answer to your question is: T2 = 235.44 °K

Explanation:

Data

V1 = 3.15 L V2 = 2.78 L

P1 = 2.40 atm P2 = 1.97 atm

T1 = 325°K T2 = ?

Formula

$\frac{P1V1}{T1} = \frac{P2V2}{T2}$

Process

T2 = (P2V2T1) / (P1V1)

T2 = (1.97x 2.78x 325) / (2.40 x 3.15)

T2 = 1779.895 / 7.56

T2 = 235.44 °K

4 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

What are these associated with the solar wind

Serjik [45]

Answer:

The solar wind is a stream of charged particles released from the upper atmosphere of the Sun, called the corona. ... Its particles can escape the Sun's gravity because of their high energy resulting from the high temperature of the corona, which in turn is a result of the coronal magnetic field.

Explanation:

5 0

3 years ago

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