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const2013 [10]
3 years ago
10

In a heating curve, when is the temperature constant? need answers!

Chemistry
2 answers:
Varvara68 [4.7K]3 years ago
3 0
During a phase change
Bingel [31]3 years ago
3 0

<u>Answer: </u>The correct answer is Option 1.

<u>Explanation:</u>

For the given options:

<u>Option 1:</u> During a phase change

Whenever phase change occurs, the temperature remains constant because the heat is used to overcome the intermolecular forces between the particles.

<u>Option 2:</u> During a pressure change

As we know pressure is directly related to the temperature. So, if the pressure is increased, the temperature will also increase and vice-versa.

<u>Option 3:</u> During a change in kinetic energy

Kinetic energy is directly related to the temperature of a substance. So, if the kinetic energy is increased, the temperature will also increase and vice-versa.

<u>Option 4:</u> During a change in molecular velocity

Molecular velocity is directly related to the temperature of a substance. So, if the molecular velocity is increased, the temperature will also increase and vice-versa.

Hence, the correct answer is Option 1.

You might be interested in
What is the specific heat of copper when heated to 221.32
Dafna1 [17]

Answer:

The specific heat of copper when heated to 221.32 (not listed form of heat measurement) is 221.32 (not listed form of heat measurement).

Explanation:

uh not really sure what else there is here, I may be missing something

5 0
2 years ago
2 Upper P Upper O Upper C l Subscript 3 Baseline (g) + heat double-headed arrow 2 Upper P Upper C l Subscript 3 Baseline (g) + U
gregori [183]

Answer:

The equilibrium will shift left.

Explanation:

Hope this helps <3

6 0
2 years ago
How many grams of KCI can be dissolved in 63.5. g of water at 80
Masteriza [31]

Answer:

35.8 g

Explanation:

Step 1: Given data

Mass of water: 63.5 g

Step 2: Calculate how many grams of KCl can be dissolved in 63.5. g of water at 80 °C

Solubility is the maximum amount of solute that can be dissolved in 100 g of solute at a specified temperature. The solubility of KCl at 80 °C is 56.3 g%g, that is, we can dissolve up to 56.3 g of KCl in 100 g of water.

63.5 g Water × 56.3 g KCl/100 g Water = 35.8 g KCl

8 0
3 years ago
The temperature of nitrogen is unknown when the gas occupies 100.0 mL at 99.10 kPa. If the gas is known to occupy 74.2 mL at 133
laiz [17]

Answer:

The unknown temperature is 304.7K

Explanation:

V1 = 100mL = 100*10^-3L

P1 = 99.10kPa = 99.10*10³Pa

V2 = 74.2mL = 74.2*10^-3L

P2 = 133.7kPa = 133.7*10³Pa

T2 = 305K

T1 = ?

From combined gas equation,

(P1 * V1) / T1 = (P2 * V2) / T2

Solving for T1,

T1 = (P1 * V1 * T2) / (P2 * V2)

T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)

T1 = 3022550 / 9920.54

T1 = 304.67K

T1 = 304.7K

3 0
3 years ago
What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of hydrogen peroxide, H2O2?
daser333 [38]

Answer:

6 mol/L

Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v          (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

50.0 mL/ 1000 mL = <em>0.05 L</em>

Now, you have to convert grams to moles in order to solve for molarity (M).

1.) On the periodic table find the molecular weights of H and O.

H= 1.01 g/mol         O= 16.00 g/mol

2.) Multiply them and then add them together to have their combined molecular weights. (You have to multiply by 2 because of their equation; H2O2).

2(1.01) + 2(16.00)= 34.02 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

10.0 g/ 34.02 g = <em>0.2939 mol</em>

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4.)You are now going to plug in your answers into the equation for Molarity.

M= 0.2939 mol / 0.05 L = <em>5.878 mol/L</em>

5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 5.878 to 6 mol/L

Sorry this explanation is very long let me know if you need a better more written out explanation.

4 0
3 years ago
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