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Katarina [22]
3 years ago
14

He particles are mostly free of each other influence is solid , liquid , gas

Chemistry
1 answer:
Simora [160]3 years ago
4 0

the most free particles are gases, then liquids, solids

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A rectangle has a length of 5.50m and a width of 12.0m. What is the area of this rectangle?​
jolli1 [7]

Answer:

66m

Explanation:

To get the area of something you multiple the length (5.5) by the width (12) together. So the problem would look like 5.5×12 and if you multipe that you get 66

3 0
3 years ago
Calculate the mass-to-mass ratio of 25.0 g of salt in 105 g of water. Work must be shown in order to earn credit.
Sveta_85 [38]

Answer:

0.23

Explanation:

  • It is known that, the mass to mass ratio of the salt to water

= (mass of salt / mass of water)

= (25.0 g / 105.0 g)

= 0.23

  • So, the answer is 0.23

4 0
3 years ago
Create a 3-D model of Bohr's atom for lithium
inn [45]

https://sciencing.com/make-3d-model-atom-5887341.html


3 0
4 years ago
The enthalpy change for converting 1.00 mol of ice at -50.0 ∘c to water at 60.0∘c is ________ kj. the specific heats of ice, wat
guajiro [1.7K]
First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

                                            = 18 g

and C is the specific heat capacity of ice = 2.09 J/g-K

and ΔT change in temperature = 0- (-50) = 50°C

by substitution:

∴q1 = 18 g * 2.09 J/g-K *50°C

       = 1881 J = 1.881 KJ

2- the heat required to melt this mass of ice is :

q2 = n*ΔHfus 

when n is the number of moles of ice = 1 mol

and ΔHfus = 6.01 KJ/mol

by substitution:

q2 = 1 mol * 6.01 KJ/mol

     = 6.01 KJ

3- the heat required to increase the water temperature from 0°C to 60 °C is:

q3 = m*C*ΔT

when m is the mass of water = 18 g 

C is the specific heat capacity of water = 4.18 J/g-K

ΔT is the change of Temperature of water = 60°C - 0°C = 60°C

by substitution:

∴q3 = 18 g * 4.18 J/g-K * 60°C

      = 4514 J = 4.514 KJ

∴the total change of enthalpy = q1+q2+q3

                                                  = 1.881 KJ  +6.01 KJ + 4.514 KJ

                                                  = 12.405 KJ


5 0
3 years ago
How many molecules are in a mole of H2O
belka [17]
1023 molecules or atoms depending on substance
7 0
3 years ago
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