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Bogdan [553]
4 years ago
8

Does all the energy travel the same way?

Chemistry
2 answers:
Sergeu [11.5K]4 years ago
5 0
No energy can travel in light, electricity, etc.
scoray [572]4 years ago
3 0
Nope.....................
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What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B
natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

A = A_{0}e^{-kt}

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

ln[A] = ln[A]_{0}-kt

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

7 0
4 years ago
Diego is trying to lift a piano to the second floor of his house. Diego uses a pulley system and gives a big lift to the piano.T
Marina86 [1]

Answer:

positive force → balanced force → negative force

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np

5 0
3 years ago
4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

8 0
3 years ago
A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)↽−−⇀H+(aq)+A−(aq) The equilibrium concentratio
Pachacha [2.7K]

Answer:

pK_{a} of HA is 6.80

Explanation:

pK_{a}=-logK_{a}

Acid dissociation constant (K_{a}) of HA is represented as-

                K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}

Where species inside third bracket represents equilibrium concentrations

Now, plug in all the given equilibrium concentration into above equation-

K_{a}=\frac{(2.00\times 10^{-4})\times (2.00\times 10^{-4})}{0.250}

So, K_{a}=1.6\times 10^{-7}

Hence pK_{a}=-log(1.6\times 10^{-7})=6.80

6 0
3 years ago
Which of the following cycles does not involve living organisms?
Aneli [31]
The correct answer is letter C. Rock Cycle. Living organisms involved in carbon cycle, oxygen cycle, and in nitrogen cycle. These are involved in the air that living organisms are taking in and out.
7 0
4 years ago
Read 2 more answers
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