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3 years ago

How does increasing the temperature affect collisions?

Chemistry

2 answers:

ddd [48]3 years ago

8 0

Combustion Have a great day!

Nastasia [14]3 years ago

8 0

Answer:

When the temperature is increased, the mean energy of the particles also increases. The frequency of collisions increases, but so does the likelihood that the collision exceeds the activation energy . Both changes mean that the proportion of collisions that are successful increases

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What is the half-life of 20 g of a radioactive sample if 5 g remain after 8 minutes?

Anni [7]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 20 g

m (final mass after time T) = 5 g

x (number of periods elapsed) = ?

P (Half-life) = ? (in minutes)

T (Elapsed time for sample reduction) = 8 minutes

Let's find the number of periods elapsed (x), let us see:

$m = \dfrac{m_o}{2^x}$

$5 = \dfrac{20}{2^x}$

$2^x = \dfrac{20}{5}$

$2^x = 4$

$2^x = 2^2$

$\boxed{x = 2}$

Now, let's find the half-life (P) of the radioactive sample, let's see:

$T = x*P$

$8 = 2*P$

$2\:P = 8$

$P = \dfrac{8}{2}$

$\boxed{\boxed{P = 4\:minutes}}\Longleftarrow(Half-Life)\end{array}}\qquad\checkmark$

I Hope this helps, greetings ... DexteR! =)

5 0

3 years ago

Tin has ten stable isotopes. The heaviest, 124Sn, makes up 5.80% of naturally occuring tin atoms. How many atoms of 124Sn are pr

matrenka [14]

The average atomic mass of Sn is 118.71 g/mol

the percentage of heaviest Sn is 5.80%

the given mass of Sn is 82g

The total moles of Sn will be = mass / atomic mass = 82/118.71=0.691

Total atoms of Sn in 82g = $6.023X10^{23}X0.691=4.16X10^{23}$

the percentage of heaviest Sn is 5.80%

So the total atoms of $Sn^{124}$ = 5.80% X $4.16X10^{23}$

Total atoms of $Sn^{124}$ = $2.41X10^{22}$ atoms

the mass of $Sn^{124}$ will be = $\frac{2.41X10^{22}X124}{6.023X10^{23}}=4.96g$

5 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$