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Tju [1.3M]
3 years ago
8

If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?

Chemistry
1 answer:
QveST [7]3 years ago
4 0
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
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Answer:

Read Explanation

Explanation:

Aluminum foil does contain billions of atoms, but we write them just as Al because aluminum does not exist in single atom forms in normal circumstances but as billions in metallic bonds.

6 0
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Could volume by displacement be used to determine the volume of a lump of rock salt? Explain would be nice
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Yes it could, but you'd have to set up the process very carefully.
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5 0
3 years ago
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7 0
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Free_Kalibri [48]

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Explanation:

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For example, vinegar (acetic acid) is used while making noodles.

Thus, we can conclude that out of the given options acetic acid is in household vinegar.

7 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
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