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Lunna [17]
3 years ago
9

A 7.11g sample of potassium chlorate was decomposed according to the following equation: 2KCLO3 -->2KCL + 3O2 How many moles

of oxygen are formed?
Chemistry
1 answer:
Ivan3 years ago
3 0

Answer:

The answer to your question is 0.087 moles of Oxygen

Explanation:

Data

mass of KClO₃ = 7.11 g

moles of O₂ = ?

Balanced chemical reaction

               2KClO₃  ⇒   2KCl  +  3O₂

Process

1.- Calculate the molar mass of KClO₃

KClO₃ = (1 x 39.1) + (1 x 35.45) + (3 x 16)

           = 39.1 + 35.45 + 48

           = 122.55 g

2.- Convert the mass to KClO₃ to moles

                  122.55 g ------------------- 1 mol

                      7.11 g   -------------------  x

                       x = (7.11 x 1)/122.55

                       x = 0.058 moles

3.- Calculate the moles of Oxygen

                  2 moles of KClO₃ ---------------- 3 moles of Oxygen

                  0.058 moles         ----------------  x

                  x = (0.058 x 3) / 2

                  x = 0.087 moles of Oxygen

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Final velocity(v) = 32 m/s
Initial velocity(u) = 10 m/s

Using kinematic equation v = u + at, 

32 = 10 + a(3)

32-10
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Hope this helps!!
8 0
3 years ago
A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate
kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of solution 1 (liquid water) = 50.0 g

m_2 = mass of solution 2 (liquid water) = 29.0 g

T_{final} = final temperature = ?

T_1 = initial temperature of solution 1 = 25°C  = [273 + 25] = 298 K

T_2 = initial temperature of solution 2 = 45°C  = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K

Converting this into degree Celsius, we use the conversion factor:

T(K)=T(^oC)+273

305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC

Hence, the final temperature of water is 32.3°C

7 0
3 years ago
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tankabanditka [31]

Answer:

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Explanation:

6 0
2 years ago
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Answer:

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