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3 years ago

7

plz help asap (plz don't give helpless answers) and also ill give brainist to those who answer the questing right

1 answer:

Lera25 [3.4K]3 years ago

5 0

1.) Particle B has a greater kinetic energy because when a particle is at higher temperature it contains more energy.

2.) Diagram B best shows the overflow of heat between the particles. It does because of the enclosed space that it is in and how it will circulate in it.

3.) Heat is transferred between object A and B by having object B (warmer) make a reaction with object A (colder).

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When one atom of sodium (Na) combines with one atom of fluorine (F), they form the compound sodium fluoride (NaF). Which of the

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Na is a cation (positive) and F is a anion (negative) this is an ionic bond. In a ionic bond the cation loses its valence electrons to give them to the anion. In this case Na will give its valence electron to F so F can complete its octet:)

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What is the oxidation number b in h3bo3?

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A Nuclear reactor produces product 20kg of uranium-235.if the half life is about 70years , how long will it take to decay to 0.1

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3 years ago

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

$t_{\frac{1}{2} }=\frac{0.693}{k}$

Here $t_{\frac{1}{2} }=0.13 min$

$k=\frac{0.693}{t_{\frac{1}{2}} }$

$\Rightarrow k=\frac{0.693}{13 }$

To find the time takes for it to decrease to 0.085 we use the below equation

$k=\frac{2.303}{t} log\frac{[A_0]}{[A]}$

$\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}$

Here , $k=\frac{0.693}{13 }$ , [A₀] = 0.13 m and [ A] = 0.085 M

$t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})$

$\Rightarrow t= 7.97\approx 8.0$

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0

3 years ago