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3 years ago

14

Define empirical formula and molecular formula with example of each

1 answer:

Arlecino [84]3 years ago

5 0

Hi -----):):$/$:); wyd n

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What is the primary result of a fission

Rudik [331]

Answer:

(1) conversion of mass to energy

Explanation:

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3 years ago

Complete the following equations

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3 years ago

Calculate the energy required to melt 21 g of ice at 0 oC.

vladimir1956 [14]

<u>We are given:</u>

Mass of ice = 21 grams

The ice is already at 0°c, the temperature at which it melts to form water

Molar heat of fusion of Ice = 6.02 kJ/mol

<u>Finding the energy required:</u>

<u>Number of moles of Ice: </u>

Molar mass of water = 18 g/mol

Number of moles = given mass/ molar mass

Number of moles = 21 / 18 = 7/6 moles

<u>Energy required to melt the given amount of ice:</u>

Energy = number of moles * molar heat of fusion

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3 years ago

Explain the kinetic energy of the particles in each of the 3 main states of matter.

Lelechka [254]

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3 years ago

Read 2 more answers

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

(c) $\Delta\mu =\Delta\mu(l)-\Delta\mu(s)$ = $-S_m\DeltaT$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago