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Lelu [443]
3 years ago
13

True or false. an object's weight is constant in all circumstances​

Chemistry
1 answer:
Sindrei [870]3 years ago
3 0

Answer: False

Explanation: False, what if its in space?

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What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?
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J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side. 

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How do i know how many neutrons of an element
Alex_Xolod [135]

Answer:

neutrons = atomic mass - atomic number.

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What makes a heterogeneous mixture different from a homogeneous mixture? The components are mixed unevenly instead of evenly wit
bazaltina [42]
<span>is a mixture that composes of components that aren't uniform or they have localized regions that all have different properties. Despite the term appearing to be highly scientific, there are various common substances that are heterogeneous mixtures

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8 0
3 years ago
For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution?
gogolik [260]
By using the ICE table :

initial    0.2 M            0           0
change -X                 + X           +X
Equ     (0.2 -X)            X               X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
∴X ≈ 10^-6
∴[HCN] = 10^-6

and PH = -㏒[H+]
             = -㏒ 10^-6
             = 6
3 0
3 years ago
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How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C
rjkz [21]

20600Cal              

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

     H  =   m  c Ф

c  = specific heat capacity of water = 4.186J/g°C

   H is the amount of heat

    Ф is the change in temperature

    H = m c (Ф₂ - Ф₁)

    H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

     1kilocalorie = 4184J

     

85996.56J to kCal; \frac{85996.56}{4184}   = 20.6kCal  = 20600Cal

               

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

6 0
3 years ago
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