Something that is bizarre fact about Radon that unique and relatively unknown by the general population??

When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit

pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

The process:

$\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq)$ .

How many moles of this process?

Relative atomic mass from a modern periodic table:

K: 39.098;
N: 14.007;
O: 15.999.

Molar mass of $\text{KNO}_3$ :

$M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of the process = Number of moles of $\text{KNO}_3$ dissolved:

$\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}$ .

What's the enthalpy change of this process?

$Q = C\cdot \Delta T = 0.505 \times (25.00 - 14.14) = 5.4843\;\text{kJ}$ for $0.212162\;\text{mol}$ . By convention, the enthalpy change $\Delta H$ measures the energy change for each mole of a process.

$\displaystyle \Delta H = \frac{Q}{n} = \frac{5.4843\text{kJ}}{0.212162\;\text{mol}} = 25.8\;\text{kJ}\cdot\text{mol}^{-1}$ .

The heat capacity is the least accurate number in these calculation. It comes with three significant figures. As a result, round the final result to three significant figures. However, make sure you keep at least one additional figure to minimize the risk of rounding errors during the calculation.

4 0

3 years ago

The following question appears on a quiz: ""You fill a tank with gas at 60Â°C to 100 kPa and seal it. You decrease the temperatu

dusya [7]

Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K$

Putting values in above equation, we get:

$\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa$

Hence, the final pressure will decrease ad the value is 85 kPa

8 0

3 years ago

Dating of Rock Layers

Sati [7]

Answer:

I'm sorry, we need the image to answer it.

Explanation:

4 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

The charge on an ion is known as its___number.

a_sh-v [17]

Answer:

oxidation number is correct!! :)

Explanation:

8 0

3 years ago