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3 years ago

8

Certain family dynamics, such as abusive parenting, can increase the susceptibility of someone developing a(n) __________ disord

er.

2 answers:

miss Akunina [59]3 years ago

6 0

Answer:

personality disorder

Explanation:

Serga [27]3 years ago

5 0

Answer:

personality disorder.

Explanation:

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two cars start at the same point and drive in a straight line for 5km. At the end of the drive their distances are the same but

Anna11 [10]

A 'displacement' always consists of a magnitude and a direction. The two cars you just described have displacements with the same magnitude ... 5 km. But if they didn't both drive in the same direction, then their displacements are different.

Remember:

-- 10 m/s² up and 10 m/s² down are different accelerations

-- 30 mph East and 30 mph West are the same speed but different velocity.

-- 5 km North and 5 km South are the same distance but different displacement.

7 0

3 years ago

In an arcade game, a 0.126 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and relea

bearhunter [10]

Answer:

4.156 m/s

Explanation:

See attachment

4 0

3 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

A certain plucked string produces a fundamental frequency of 150 hz. Which frequency is not one of the harmonics produced by tha

Irina-Kira [14]

Answer:

Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...

Explanation:

Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.

For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc

If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc

Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...

6 0

3 years ago

If an electron is accelerated from rest through a potential difference of 1200V find its approximate velocity at the end of this

kolbaska11 [484]

Answer: 2.1 × 10^7 m/s

Explanation:

Please see the attachments below

8 0

4 years ago