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3 years ago

13

Which of the following is a transuranium element? Ra Am Tc Pa

1 answer:

lyudmila [28]3 years ago

3 0

Am - it has an atomic number of 95 which is greater than 92.

Transuranium elements are elements with atomic levels greater than 92

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Una sustancia tiene una masa de 81632 g. Se logró desplazar aplicando un trabajo de 2800 J. Averigüe la distancia que logró move

Rudik [331]

Answer:

Supongo que su suma 81632 + 2800 = 84432, por lo que la distancia que logró mover el objeto fue 2800.

Explanation:

7 0

3 years ago

The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be

Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature $T_1 =$ 1001 K

Final temperature $T_2 =$ 298 K

Applying the equation of Arrhenius theory.

$In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})$

where ;

R gas constant = 8.314 J/K/mol

$In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})$

$In \dfrac{k_2}{0.380 }= -70.8655$

$\dfrac{k_2}{0.380 }= e^{-70.8655}$

$\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}$

${k_2}= 1.67303256 \times 10^{-31} \times {0.380 }$

${k_2}= 6.3575 \times 10^{-32}$ /M .sec

Half life:

At 1001 K.

$t_{1/2} = \dfrac{In_2}{k_1}$

$t_{1/2} = \dfrac{0.693}{0.38}$

$t_{1/2} =$ 1.82368 secc

At 298 K:

$t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}$

$t_{1/2} =1.09 \times 10^{31} \ sec$

3 0

3 years ago

For the following hypothetical reaction:

olga55 [171]

The moles of B that will be needed to convert 2 moles of A into as many moles of C as possible is 6 moles

Explanation

3A +9B → 5C

The moles of B are calculated using the mole ratio.

That is; from the equation above the mole ratio of A:B is 3:9

If the moles of A required is 2 moles therefore the moles of B

= 2 x9/3= 6 moles

3 0

3 years ago

A quantity of 8.10 × 102 mL of 0.600 M HNO3 is mixed with 8.10 × 102 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of

victus00 [196]

Answer:

22.48°C is the final temperature of the solution.

Explanation:

Heat of neutralization of reaction , when 1 mol of nitric acid reacts= ΔH= -56.2 kJ/mol= -56200 J/mol

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of nitric acid = n

Volume of nitric acid solution = $8.10\times 10^2 mL= 0.81L$

Molarity of the nitric acid = 0.600 M

$n=0.600 M\times 0.81 L=0.486 mol$

Moles of barium hydroxide = n'

Volume of barium hydroxide solution = $8.10\times 10^2 mL= 0.81L$

Molarity of the barium hydroxide= 0.300 M

$n'=0.300 M\times 0.81 L=0.243 mol$

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_3+2H_2O$

According to reaction, 2 mol of nitric acid reacts with 1 mol of barium hydroxide .Then 0.486 mol of nitric acid will react with :

$\frac{1}{2}\times 0.486 mol=0.243 mol$ barium hydroxide.

Heat release when 0.486 mol of nitric acid reacted = Q

= ΔH × 0.486 = -56200 J/mol × 0.486 mol=-27,313.2 J

Heat absorbed by the mixture after reaction = Q' = -Q = 27,313.2 J

Volume of the nitric solution = 0.81 L = 810 mL

Volume of the ferric nitrate solution = 0.81 L = 810 mL

Total volume of the solution = 810 mL + 810 mL = 1620 mL

Mass of the final solution = m

Density of water = density of the final solution = d = 1 g/mL

$Mass=density\times Volume$

$m=1 g/ml\times 1620 ml=1620 g$

Initial temperature of the both solution were same = $T_1=18.46^oC$

Final temperature of the both solution will also be same after mixing= $T_2$

Heat capacity of the mixture = c = 4.184 J/g°C

Change in temperature of the mixture = ΔT = $(T_2-T_1)$

$Q=mc\Delta T=mc(T_2-T_1)$

$27,313.2 J= 1620 g\times 4.184 J/g^oC\times (T_2-18.46^oC)$

$T_2=22.49^oC$

22.48°C is the final temperature of the solution.

8 0

3 years ago

On a hot summer day, a piece of ice is removed from a cooler and placed in a cup that is sitting on a picnic table. Which of the

dlinn [17]

The ice absorbs heat from the cup . And the ice loses its composition thus melting .

5 0

3 years ago