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tino4ka555 [31]
3 years ago
11

Which of the following quantities can be determined from a speed-time graph of a particle travelling in a straight line?

Physics
2 answers:
enot [183]3 years ago
8 0

The answer is:

Both the distance traveled in a given time and the magnitude of the acceleration at a given instant


Hope I Helped!

Diano4ka-milaya [45]3 years ago
7 0
The answer is letter B
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In the United States, where can volcanoes be found?
Sedbober [7]
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7 0
2 years ago
ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami
yuradex [85]

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

Explanation :

The given balanced chemical reaction is,

2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)

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

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

\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]

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=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]

\Delta H^o=62.2kJ=62200J

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_{product}

\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]

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=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]

\Delta S^o=132.67J/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 298 K.

\Delta G^o=(62200J)-(298K\times 132.67J/K)

\Delta G^o=22664.34J=22.66kJ

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

6 0
3 years ago
The model below shows a calcium atom. An image has a mix of red and blue balls in its center and 4 concentric black rings around
zubka84 [21]

Answer:

8 electrons in the third energy level

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From the description,the third energy level has 8 electron (represented by the small green balls you describe)

4 0
3 years ago
The period T of a pendulum of length L is measured to determine g at the surface of Earth. The equation used is T=2π√L/g. The ma
saul85 [17]

Answer:

C: Variation in the value of g as the pendulum bob moves along its arc.

Explanation:

The formula for period of a simple pendulum is given by;

T = 2π√(L/g)

Where;

L is length

g is acceleration due to gravity

Now, from this period equation, it is clear that the only thing that can affect the period of a simple pendulum are changes to its length and acceleration due to gravity.

Looking at the options, the only one that talks about either the length or gravity as being potential causes of the error is option C

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One of the factors that determines the ? of a capacitor is the frequency measured in hertz.
ycow [4]

Answer:

Reactance

Explanation:

In an AC circuit, the capacitive reactance of a capacitor is given by:

X_C = \frac{1}{2 \pi f C}

where

f is the frequency of the AC current

C is the capacitance of the capacitor

The reactance of the capacitor tells somehow the "resistance" of the capacitor to the passage of current through it. In fact:

- When the frequency of the AC current is zero (this means, we are in regime of DC current), the reactance becomes infinite, and this is true because the capacitor does not let the current pass through it)

- When the frequency of the AC current tends to infinite, the reactance becomes zero, and this is true because in this case the current changes direction so fast that the capacitor has not enough time to "block" the current, so the current almost no feels the presence of the capacitor.

7 0
3 years ago
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