Determine the mass in grams of 125 mol of neon

Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a

JulijaS [17]

Answer : The correct option is, $30.9^oC$

Explanation :

Formula used :

$q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

where,

$q$ = heat released = 24 KJ

$m$ = mass of bomb calorimeter = 1.30 Kg

$c$ = specific heat = $3.41J/g^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25.5^oC$

Now put all the given values in the above formula, we get the final temperature of the calorimeter.

$q=m\times c\times (T_{final}-T_{initial})$

$24KJ=1.30Kg\times 3.41J/g^oC\times (T_{final}-25.5)^oC$

$T_{final}=30.9^oC$

Therefore, the final temperature of the calorimeter is, $30.9^oC$

5 0

2 years ago

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A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

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

$Q_{water} = Q_{sample}$

Where:

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

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

7 0

2 years ago

The solubility of glucose at 30°C is 125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water a

seraphim [82]

The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g.

The concentration of the solution in water is,
550 g/400g of water = 1.375 g glucose / g of water

Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.

4 0

2 years ago

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In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as

lord [1]

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio of any nuclei ranges from to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons = Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons = Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons = Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons = Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons = Neutron- proton ratio = N/P = 7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons = Neutron- proton ratio = N/P = 0/1 =0 = unstable

6 0

2 years ago

Bobby's mom sets a small pot of water on the stove and lights the burner. Ten minutes later, Bobby notices small bubbles and see

dolphi86 [110]

There are no answers, list them so I can answer.

7 0

1 year ago

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