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lesya692 [45]
3 years ago
6

A thermometer is placed in a beaker full of water that has been sitting on the lab table for a few hours. The beaker is placed o

n a ring stand and heated over a Bunsen burner for 4 minutes. After 4 minutes, the temperature is taken again. Find the temperature change of the water if the temperature of the room is 26°C.
Chemistry
1 answer:
meriva3 years ago
7 0
The temperature would be 30C if the water is at the same temp as the lab  <span />
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On a recent trip to the Fort Worth Zoo, Trevor's parents bought him a helium-filled balloon. At the beginning of the day, the sk
timurjin [86]

Answer:

1. The new volume is 14L

2. The volume decreased

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 14.2 L

Initial pressure (P1) = 102.5 kPa

Initial temperature (T1) = 33°C

Final pressure (P2) = 100.9 kPa

Final temperature (T2) = 24°C.

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

K = °C + 273

T1 = 33°C = 33°C + 273 = 306K

T2 = 24°C = 24°C + 273 = 297K

Step 3:

Determination of the new volume.

Applying the general gas equation P1V1/T1 = P2V2/T2, the new volume can be obtain as follow:

P1V1/T1 = P2V2/T2

102.5 x 14.2/306 = 100.9 x V2/297

Cross multiply to express in linear form as shown below:

306 x 100.9 x V2 = 102.5 x14.2 x297

Divide both side by 306 x 100.9

V2 = (102.5x14.2x297)/(306 x 100.9)

V2 = 14 L

The new volume is 14 L.

Step 4:

Determination of the change in volume. This is illustrated below

Final volume (V2) = 14 L

Initial volume (V1) = 14.2 L

Change in volume (ΔV) =?

Change in volume (ΔV) = Final volume (V2) - Initial volume (V1)

ΔV = V2 - V1

ΔV = 14 - 14.2

ΔV = - 0.2L

Since the change in volume is negative, it means there is a decrease in the volume.

7 0
3 years ago
Read 2 more answers
Is diphenylamine a solid, liquid, or a gas at room temperature
goblinko [34]

Answer:

Solid

Explanation:

Diphenylamine has a melting point of 127.4 F or 53 C so at room temperature ~70 F or 21 C its a solid.

3 0
3 years ago
An aqueous solution of glucose has molality of 6.81 and a density of 1.2 g/ml what is the molarity of glucose in the solution
andriy [413]

Answer:

the molarity is 3.68 moles/L

Explanation:

the molality of the solution of sucrose is

m= moles of glucose / Kg of solvent (water)= 6.81 ,

since the molecular weight of glucose is 180.156 gr/mole , then per each kilogram of solvent there is

6.81 moles*180.156 gr/mole + 1000 gr of water = 2226.86 gr of solution

from the density

volume of solution = mass of solution/density = 2286.86 gr / 1.2 gr/ml = 1855.71 ml

therefore there is 1000 gr of water in 1855.71 ml

then the molarity M is

M= moles of glucose / L of solution = (moles of glucose /  Kg of solvent) *  (Kg of solvent/L of solution) = 6.81 moles/Kg * 1Kg/1.85 L = 3.68 moles/L

M= 3.68 moles/L

Note:

- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.

3 0
3 years ago
Write a balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air.
Lelechka [254]

Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O.

Explanation:

When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O

Number of atoms present on reactant side are as follows.

  • C = 1
  • H = 4
  • O = 2

Number of atoms present on product side are as follows.

  • C = 1
  • H = 2
  • O = 3

To balance this equation, multiply O_{2} by 2 on reactant side. Also, multiply H_{2}O by 2 on product side. Hence, the equation can be rewritten as follows.

CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O

Now, the number of atoms present on reactant side are as follows.

  • C = 1
  • H = 4
  • O = 4

Number of atoms present on product side are as follows.

  • C = 1
  • H = 4
  • O = 4

Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

Thus, we can conclude that balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O.

4 0
3 years ago
How could you safely determine if a base is stronger than an acid?
expeople1 [14]
You should never taste anything in a lab.
Litmus paper will only tell you whether it is an acid or a base, not how strong it is a hydrogen test only works with acids.
You can get a serious burn from touching strong acids and bases and it doesn't give an objective measure of the strength.
The only one that is left is conductivity. This is safe and rather reliable, because the "strength" of an acid or a base depends on how much it dissociates. Hence more dissociation means greater conductivity.
6 0
3 years ago
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