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Tamiku [17]
4 years ago
8

How does the amount of salt dissolve in water affect the density of water

Chemistry
1 answer:
ad-work [718]4 years ago
3 0

Answer:

Adding salt to the water increases the density of the solution because the salt increases the mass without changing the volume very much. When you add table salt (sodium chloride, NaCl) to water, the salt dissolves into ions, Na+ and Cl-. The volume increases by a small factor, but the mass increases by a bigger factor.

Explanation:

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Difference between atom and molecule​
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Atoms are single neutral particles. Molecules are neutral particles made of two or more atoms bonded together. An ion is a positively or negatively charged particle.

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The pressure of a gas is caused by
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The rapid motion and collisions of molecules with the walls of the container causes pressure.

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What is an endothermic reaction?
GrogVix [38]
C is the right answer as endothermic reaction takes in the heat. That is why cooling effect is observed whenever such reaction takes place.
Hope this helped :)
3 0
3 years ago
In calculating the concentration of [Cu(NH3)4]2+ from [Cu(H2O)4]2+, the stepwise formation constants are as follows: K1=1.90×104
pashok25 [27]

Answer:

Kf = 1.11x10¹³

Explanation:

The value of Kf for a multistep process that involves an equilibrium at each step, is the multiplication of the constant of the equilibrium of each step.

Kf = K1xK2xK3xK4

Kf = 1.90x10⁴ x 3.90x10³ x 1.00x10³ x 1.50x10²

Kf = 1.11x10¹³

4 0
3 years ago
How much work must be done on a 5-kg snowboard to increase its speed from 2 m/s to 4 m/s?
Alekssandra [29.7K]

Answer:

Option C = 30 j

Explanation:

Given data:

mass of snowboard = 5 Kg

Initial speed = 2 m/s

final speed = 4 m/s

work done = ΔE= ?

ΔE= change in kinetic energy

Solution:

Formula:

K.E (initial) = 1/2 × mv²

K.E (initial)  = 1/2 × 5 Kg . (2m/s)²

K.E (initial) = 1/2 × 20 Kg.m²/s²

K.E (initial) = 10 Kg.m²/s² or 10 J      

   Kg.m²/s² = J

K.E (finial) = 1/2 × mv²

K.E (finial) = 1/2 ×  5 Kg . (4m/s)²

K.E (finial) = 1/2 ×  5 Kg . 16 m²/s²

K.E (finial) = 1/2 × 80 Kg.m²/s²

K.E (finial) = 40 Kg.m²/s²  or 40 J

work done =  ΔE = K.E (finial) - K.E (initial)

work done =  ΔE = 40 J - 10 J

work done =  ΔE = 30 J

6 0
3 years ago
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