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A 237g sample of molybdnum metal is heated to 100.1 0C and then dropped into an insulated cup containing 244 g of water at 10.0

miv72 [106K]

Answer:

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

Explanation:

We consider the system formed by the molybdenum metal and water as our system, a control mass inside an insulated cup, that is, a container that avoids any energy and mass interactions between system and surroundings.

From statement we notice that metal is cooled down whereas water is heated. According to the First Law of Thermodynamics, we know that:

$Q_{metal} - Q_{water} = 0$

$Q_{metal} = Q_{water}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{metal}$ - Heat released by metal, measured in joules.

Now we expand this identity by definition of sensible heat:

$m_{metal}\cdot c_{metal}\cdot (T_{m,o}-T) = m_{water}\cdot c_{water}\cdot (T-T_{w,o})$

The specific heat of the metal is cleared within equation above:

$c_{metal} = \frac{m_{water}\cdot c_{water}\cdot (T-T_{w,o})}{m_{metal}\cdot (T_{m,o}-T)}$

If we know that $m_{water} = 0.237\,kg$ , $m_{metal} = 0.244\,kg$ , $c_{water} = 4186\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{w,o} = 10\,^{\circ}C$ , $T_{m,o} = 100.10\,^{\circ}C$ and $T = 15.30\,^{\circ}C$ , the specific heat of molybdenum is:

$c_{metal} = \frac{(0.237\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (15.30\,^{\circ}C-10\,^{\circ}C)}{(0.244\,kg)\cdot (100.10\,^{\circ}C-15.30\,^{\circ}C)}$

$c_{metal} = 254.119\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of molybdenum is 0.254 joules per gram-Celsius.

5 0

3 years ago

What is the Na+ concentration in each of the following solutions:

iren [92.7K]

Sodium Sulfate = Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M stands for Molarity, defined as Molarity = (moles of solute)/(Liters of solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65 moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+ ions in solution for every one Na2SO4.

Therefore if 3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and since this is still a theoretical solution, we can assume 1 L of solution.

Finally we find [Na+] = 2*3.65 = 7.3M

Use the same logic for parts b and c

5 0

3 years ago

A car with a mass of 1500 kg is moving at a rate of 4 m s . Determine its kinetic energy

rodikova [14]

the correct answer is b) 12000 J

4 0

3 years ago

Read 2 more answers

According to the image, identify the number of neutrons in the most common isotope of aluminum.

wolverine [178]

Answer:

The number of neutron in the Aluminium Isotope is :

B. 14

Explanation:

Isotopes : These are the atoms which have same atomic number but have different mass number.

<u>This image shows the average atomic mass of Al element because it is in decimals</u>.

Atomic mass = 26.98154

(Note : mass number of single isotope can never be in decimals)

It is the average of mass of different isotopes of Al

Major Isotopes of $_{13}^{26.98154}\textrm{Al}$ are :

$_{13}^{26}\textrm{Al}$ ......atomic mass = 26
$_{13}^{27}\textrm{Al}$ .......atomic mass = 27

mass of Al given in image(26.98) is nearly equal to mass of 2nd isotope(27)

mass of $_{13}^{26.98154}\textrm{Al}\ \approx 27$

Now calculate the neutron in $_{13}^{27}\textrm{Al}$

Number of neutron = mass number - atomic number

= 27 - 13

Number of neutron = 14

(Atomic mass is same as mass number)

5 0

3 years ago

What is the necessary voltage to power the electrolysis of molten sodium chloride?

sammy [17]

4V is the necessary voltage to power the electrolysis of molten sodium chloride.

To create sodium metal and chlorine gas, molten (liquid) sodium chloride can be electrolyzed. A Down's cell is the name of the electrolytic cell utilised in the procedure. The liquid sodium ions in a Down's cell are converted to liquid sodium metal at the cathode. Liquid chlorine ions are oxidised to chlorine gas at the anode. Below is an illustration of the reactions and cell potentials:

oxidation: $2Cl^{-} (l)$ → $Cl_{2} (g)$ + $2e^{-}$ E°= -1.36V

reduction: $Na^{+} (l) + e^{-}$ → $Na (l)$ E°= -2.71V

overall : $2Na^{+} (l) + 2Cl^{-} (l)$ → $2Na(l) +Cl_{2} (g)$ E° $_{cell}$ = -4.07V

For this electrolysis to take place, the battery needs to supply more than 4 volts. The only means to obtain pure sodium metal is by this reaction, which also serves as a significant source of chlorine gas generation. Swimming pools and other surfaces are frequently cleaned and disinfected with chlorine gas.

Learn more about sodium chloride here;

brainly.com/question/9811771

#SPJ4

6 0

2 years ago