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makkiz [27]
3 years ago
6

Describe what happens to water particles when ice melts

Chemistry
1 answer:
Rama09 [41]3 years ago
4 0
Hello there! I'm gladly to help you out! :)

Melting Ice melts when heat energy causes the molecules to move faster, breaking the hydrogen bonds between molecules to form liquid water.

Hope this helped!
You might be interested in
Need help on last 3 questions
Olin [163]

Answer:

Explanation:

1)

Given data:

Initial volume of balloon = 0.8 L

Initial temperature = 12°C ( 12+273= 285 K)

Final temperature = 300°C (300+273 = 573 K)

Final volume = ?

Solution:

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 0.8 L .573 K / 285 K

V₂ = 458.4 L / 285

V₂ = 1.61 L

2)

Initial pressure = 204 kpa

Initial temperature = 29°C ( 29 + 273 = 302 K)

Final temperature = ?

Final pressure = 300 kpa

Solution:

P₁/T₁ = P₂/T₂

T₂ = T₁P₂/P₁  

T₂ = 302 K . 300 kpa / 204 kpa

T₂ = 90600 K/ 204

T₂ = 444.12 K

3)

Given data:

Initial volume = 14 L

Initial pressure = 2.1 atm

Initial temperature = 100 K

Final temperature = 450 K

Final volume = ?

Final pressure = 1.2 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm  

V₂ = 13230 L / 120

V₂ = 110.25 L

5 0
3 years ago
HELP QUICK!! Match the term with the definition. (4 points)
Rufina [12.5K]

Answer:

1. c.

2. d.

3. b.

4. a.

Explanation:

7 0
3 years ago
Read 2 more answers
A gas occupies the volume of 215ml at 15C and 86.4kPa?
kherson [118]

Answer:

About 0.1738 liters

Explanation:

Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

0.8527\cdot 0.215=n \cdot 0.0821 \cdot 288\\n\approx 7.754 \cdot 10^{-3}

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

1V=7.754 \cdot 10^{-3} \cdot 0.0821 \cdot 273\\V\approx 0.1738L

Hope this helps!

3 0
3 years ago
What is the volume of 3.5 moles of oxygen gas (O2) at standard temperature and pressure (STP)?
baherus [9]
1 mole ----------- 22.4 L ( at STP )
3.5 mols --------- ?

V  = 3.5 x 22.4 / 1

V = 78.4 L

hope this helps!


4 0
3 years ago
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

3 0
3 years ago
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