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Alexeev081 [22]
3 years ago
12

Density is not a physical property of mass. True False

Chemistry
2 answers:
Yakvenalex [24]3 years ago
7 0
I think it's false but don't be mad if it's not right
SIZIF [17.4K]3 years ago
6 0
True it is not a physical property of mass .
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How can you find the number of neutrons that are in an atom of an element
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You just have to take the periodic table of the elements, take the element that interests you. Then, you look at the formula of the element, at the top left you find the number of neutrons and protons and at the bottom left you find the number of protons. Then you just have to make the number from the top left minus the number from the bottom left.

Explanation:

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Bubbles are released when nitric acid is added to a potassium carbonate solution.
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2 years ago
The length breadth and thickness of a brick is 18cm 8 cm and 5cm respictively find the area of the widest part of rhe brick​
Sunny_sXe [5.5K]

Answer:

144cm²

Explanation:

Given dimensions:

     Length x breadth x thickness

        18cm  x   8cm    x    5cm

The widest part of this figure will be the face containing the length and the breadth.

The breadth is the width of the figure;

Area of the widest part  = length x breadth = 18cm x 8cm  = 144cm

The area of the widest part of the figure is 144cm²

4 0
3 years ago
Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

7 0
3 years ago
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