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2 years ago

15

13 POINTS PLEASE HELP ASAP IM ON A TIME LIMIT ;-; WILL MARK BRAINLIEST what is the density of a sample that has a volume of 8c

m cubed and a mass of 45g? show all steps

2 answers:

galina1969 [7]2 years ago

7 0

Answer:

5.63

Explanation:

Divide mass by volume, which will give you 5.63. For instance, 45 divide by 8 what is it? find out

madreJ [45]2 years ago

3 0

Density = (mass) divided by (volume)

I sure hope you're not taking a test. Brainly doesn't allow questions from quizzes, exams, or tests.

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Changing the speed of a synchronous generator changes A) the frequency and amplitude of the output voltage. B) only the frequenc

Dmitrij [34]

Answer:

A) the frequency and amplitude of the output voltag

Explanation:

Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.

Changing the speed regulator will change the engine throttle setting to maintain the speed.

While the power, torque, current, fuel flow rate and torque angle will have decreased.

8 0

3 years ago

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:

Westkost [7]

Answer:

dorsiflexion

Explanation:

To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion

7 0

2 years ago

Read 2 more answers

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

When operated on a household 110.0 V line, typical hair dryers draw about 1450 W of power. The current can be modeled as a long,

dezoksy [38]

Answer:

0.075A

Explanation:

We can consider this system as a circuit, hence we can take the current from the formula for the electric power as follow

$P=IV\\I=\frac{P}{V}=\frac{110V}{1450W}=0.075A$

I hope this is useful for you

regards

4 0

2 years ago

What’s the units of specific heat

DiKsa [7]

Hello Esmeralda

Specific heat is measured in Joules per g times degree Celsius.

I hope this helps!

6 0

3 years ago