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nlexa [21]
3 years ago
12

when you put a book on a table the table pushes on the book is that newton's 1st,2nd,or 3rd law of motion ? 

Chemistry
1 answer:
ohaa [14]3 years ago
8 0
The table pushes the book and the book pushes the table
It's 3rd law
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The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the
IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

Explanation :

The expression for first order reaction is:

[C_t]=[C_o]e^{-kt}

where,

[C_t] = concentration at time 't'  (final) = ?

[C_o] = concentration at time '0' (initial) = 0.100 M

k = rate constant = 5.40\times 10^{-3}s^{-1}

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}

[C_t]=4.05\times 10^{-4}M

Thus, the concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

4 0
3 years ago
A 25.0 g bold made of an alloy absorbed 250 J of heat as its temperature changed from 25.0 °C to 78.0 °C. What is the specific h
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Answer:

Specific heat of alloy = 0.2 j/ g.°C

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

Mass of bold = 25 g

Heat absorbed = 250 J

Initial Temperature = 25°C

Final temperature = 78°C

Specific heat of alloy = ?

Solution:

Change in temperature:

ΔT = 78°C - 25°C

ΔT = 53°C

Now we will put the values in formula.

Q = m.c. ΔT

250 j = 25 g × c ×53°C

250 j = 1325 g.°C × c

250 j / 1325 g.°C = c

c = 0.2 j/ g.°C

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