20 mL of 80°C water is mixed with 20 mL of 0°C water in a perfect calorimeter. What is the final temperature?

Which of these elements has the smallest atomic radius?

Elodia [21]

Krypton. The atomic radius decreases as you go across a period.

5 0

3 years ago

What is the difference between energy levels, sub levels, and orbitals?

Lorico [155]

Explanation:

Orbitals are spaces that have a high probability of containing an electron. ... The s sublevel has just one orbital, so can contain 2 electrons max. The p sublevel has 3 orbitals, so can contain 6 electrons max. The d sublevel has 5 orbitals, so can contain 10 electrons max.

7 0

3 years ago

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

Which of the following statements is true? A) This reaction will be spontaneous only at high temperatures. B) This reaction will

viktelen [127]

Answer:

D) This reaction will be nonspontaneous only at high temperatures.

Explanation:

According the equation of Gibb's free energy -

∆G = ∆H -T∆S

∆G = is the change in gibb's free energy

∆H = is the change in enthalpy

T = temperature

∆S = is the change in entropy .

And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

ΔG < 0 , the reaction is Spontaneous

ΔG > 0 , the reaction is non Spontaneous

ΔG = 0 , the reaction is at equilibrium

The reaction has the value for ∆H = negative , and ∆S = negative ,

Now ,

∆G = ∆H -T∆S

= ( - ∆H ) - T( - ∆S )

= ( - ∆H ) +T( ∆S )

Now, for making the reaction Spontaneous ΔG = negative ,

Hence ,

The temperature is low, then the value for ΔG will be negative , i.e. , Spontaneous reaction .

And , vice versa , at higher temperature , the reaction will have ΔG positive , and the reaction will be non -Spontaneous reaction .

The standard free energy of formation will be zero , only for the compounds that are in their pure form ,

Hence , Al(s) will have ΔG = 0 .

7 0

3 years ago

A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th

Kisachek [45]

Answer:

$pH=4.1$

Explanation:

Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

$pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1$

Regards.

3 0

3 years ago