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2 years ago

Which statement about spontaneous and non spontaneous processes is correct

1 answer:

8 0

Spontaneous processes are processes that occurs freely they do not need any other external factors to be supplied for them to proceed while non-spontaneous ones requires input of energy for it to happen. Example of a spontaneous process is the decay of a diamond to graphite while the reverse of this is a non spontaneous process since heat and pressure are to be supplied for graphite to turn into diamond.

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A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How

IrinaK [193]

Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $783J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 254 g

$\Delta T$ = change in temperature = $T_2-T_1=(23.73-26.01)=-2.28^oC$

Now put all the given values in the above formula, we get:

$q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]$

$q=-4208.28J=-4.81kJ$

Therefore, the amount of heat evolved by a reaction is, 4.81 kJ

7 0

3 years ago

Why do nitrogen and oxygen form negative ions, not positive ones, in simple binary compounds.

gavmur [86]

Answer:

it is easier for them to have an octet of electrons(8e)

Each ion will: obtain the noble gas structure, each atom has high ionization energy

Explanation:

5 0

2 years ago

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$