How many mg is 59.0 kg in scientific<br> notation?

When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.643 ∘C . When a 0.275-g sample

andrew-mc [135]

Answer : The heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

Explanation :

First we have to calculate the specific heat calorimeter.

Formula used :

$Q=m\times c\times \Delta T$

where,

Q = heat of combustion of benzoic acid = 26.38 kJ/g = 26380 J/g

m = mass of benzoic acid = 0.235 g

c = specific heat of calorimeter = ?

$\Delta T$ = change in temperature = $1.643^oC$

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

$26380J/g=0.235g\times c\times 1.643^oC$

$c=68323.38J/^oC$

Thus, the specific heat of calorimeter is $68323.38J/^oC$

Now we have to calculate the heat of combustion of caffeine.

Formula used :

$Q=c\times \Delta T$

where,

Q = heat of combustion of caffeine = ?

c = specific heat of calorimeter = $68323.38J/^oC$

$\Delta T$ = change in temperature = $1.584^oC$

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

$Q=68323.38J/^oC\times 1.584^oC$

$Q=108224.23J=108.2kJ$

Now we have to calculate the moles of caffeine.

$\text{Moles of caffeine}=\frac{\text{Mass of caffeine}}{\text{Molar mass of caffeine}}$

Mass of caffeine = 0.275 g

Molar mass of caffeine = 194.19 g/mole

$\text{Moles of caffeine}=\frac{0.275g}{194.19g/mole}=0.00142mol$

Now we have to calculate the heat of combustion per mole of caffeine at constant volume.

$\text{Heat of combustion per mole of caffeine}=\frac{108.2kJ}{0.00142mol}=76197.18kJ/mole$

Therefore, the heat of combustion per mole of caffeine at constant volume is 76197.18 kJ/mole

4 0

3 years ago

Science / Chemistry Worksheet

kaheart [24]

Explanation:

K2O Now the cation is the element at the front so it would be K2 because the 2 belongs with the K which is Potassium and now Anion is the last element O for oxygen

7 0

4 years ago

What is the Scientific<br> Method?

Doss [256]

Answer:

I just finished a unit on Scientific Method in my science class! Anyway, it's defined in the screenshots below. Hope this helps!

8 0

3 years ago

Suppose that you use a pair of chopsticks and apply a force of 1 N over a distance of 0.01 m. How much work do you do? If the ou

Leto [7]

Explanation:

Work done = force * perpendicular distance

= 1 * 0.01 = 0.01 joules

7 0

3 years ago

The standard free energy of activation of a reaction A is 81.9 kJ mol–1 (19.6 kcal mol–1) at 298 K. Reaction B is one million ti

klasskru [66]

Answer:

54.9 kJ/mol

Explanation:

The relation between the activation energy (Ea) and the rate constant (k) is given by the Arrhenius equation.

$k=A.e^{-Ea/RT}$

where,

A is a collision factor

R is the ideal gas constant

T is the absolute temperature

Reaction B is one million times faster than reaction A at the same temperature. So $k_{B}=10^{6} k_{A}$ .

Then,

$k_{B}=10^{6} k_{A}\\A.e^{-Ea_{B}/RT}=10^{6}A.e^{-Ea_{A}/RT}\\e^{-Ea_{B}/RT}=10^{6}e^{-Ea_{A}/RT}\\ln(e^{-Ea_{B}/RT})=ln(10^{6}e^{-Ea_{A}/RT})\\\frac{-Ea_{B}}{RT} =ln10^{6} -\frac{Ea_{A}}{RT} \\Ea_{B}=(ln10^{6} -\frac{Ea_{A}}{RT}).(-RT)=(ln10^{6}-\frac{89.1kJ/mol}{(8.314\times 10^{-3} kJ/mol.K).298K} ).(-8.314\times 10^{-3} \frac{kJ}{mol.K}.298K )=54.9kJ/mol$

8 0

3 years ago

How many mg is 59.0 kg in scientific notation?