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mr_godi [17]
2 years ago
9

What happened to the volume of gas when the syringe was exposed to various temperature conditions? Using the concepts explored i

n the Introduction, describe why this occurred, keeping in mind the definition of temperature.
Chemistry
1 answer:
rewona [7]2 years ago
5 0

Answer:

Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.

Explanation:

The expression for Charles's Law is shown below:

\frac {V_1}{T_1}=\frac {V_2}{T_2}

This states that the volume of the gas is directly proportional to the absolute temperature keeping the pressure conditions and the moles of the gas constant.

<u>Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature. </u>

<u>For example , if the temperature of the gas is reduced to half, the volume also reduced to half. </u>

<u>At -273.15 K, according to Charles's law, it is possible to make the volume of an ideal gas = 0.</u>

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an unknown metal of mass 512 g at a temperature of 15C is dropped into 325 g of water held in a 100 g aluminum container at the
ziro4ka [17]

Answer:

The specific heat capacity of the metal is 0.843J/g°C

Explanation:

Hello,

To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.

Heat gained by the metal = heat loss by water + calorimeter

Data,

Mass of metal (M1) = 512g

Mass of water (M2) = 325g

Initial temperature of the metal (T1) = 15°C

Initial temperature of water (T2) = 98°C

Final temperature of the mixture (T3) = 78°C

Specific heat capacity of metal (C1) = ?

Specific heat capacity of water (C2) = 4.184J/g°C

Heat loss = heat gain

M2C2(T2 - T3) = M1C1(T3 - T1)

325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)

1359.8 × 20 = 512C1 × 63

27196 = 32256C1

C1 = 27196 / 32256

C1 = 0.843J/g°C

The specific heat capacity of the metal is 0.843J/g°C

8 0
2 years ago
A 6.0M solution HCl is diluted to 1.0M How many milliliters of the 6.0M solution would be used to prepare 100.o mL of the dilute
Margarita [4]
Wats r is the most soluble object ever. It is dense
4 0
2 years ago
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If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln
Ksju [112]
Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
5 0
2 years ago
Read 2 more answers
What type of bond would form between two atoms of selenium?
siniylev [52]

Answer:

the correct answer is B

hope it helps !!!

8 0
2 years ago
When an aldose reacts with Barfoed's reagent, what type of organic compound forms? What type of chemical is this?
Fudgin [204]
Barfoed's test is a concoction test utilized for identifying the nearness of monosaccharides. It depends on the diminishment of copper(II) acetic acid derivation to copper(I) oxide (Cu2O), which frames a block red hasten. 
Barfoed's reagent comprises of a 0.33 molar arrangement of unbiased copper acetic acid derivation in 1% acidic corrosive arrangement. The reagent does not keep well and it is, thusly, fitting to make it up when it is really required. May store uncertainly as per a few MSDS's.
7 0
2 years ago
Read 2 more answers
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