Search...

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

Leto [7]

Answer: The nuclide symbol of X is $^{231}_{90}\textrm{Th}$

Explanation:

The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.

The general equation representing alpha decay process is:

$_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\textrm{He}$

For the given equation :

$^{235}_{92}\textrm{U}\rightarrow ^{A}_{Z}\textrm{X}+^4_2\textrm{He}$

As the atomic number and mass number must be equal on both sides of the nuclear equation:

$^{235}_{92}\textrm{U}\rightarrow ^{231}_{90}\textrm{Th}+^4_2\textrm{He}$

Thus the nuclide symbol of X is $^{231}_{90}\textrm{Th}$

5 0

2 years ago

Which of the following occurs in this system?

earnstyle [38]

Answer:

Hence option C is correct.

Explanation:

the process is osmosis a process by which molecules of a solvent tend to pass through a semipermeable membrane from a less concentrated solution into a more concentrated one.

so the answer for the this question will be There is a net flow of water from the 4% starch solution into the 10% starch solution because solvent (water) moving from low conc. (4 %) to high conc. (10%)

Hence option C is correct.

5 0

2 years ago

The gas cyclobutane, C4H8(g), can be used in welding. When cyclobutane is burned in oxygen, the reaction is: C4H8(g) + 6 O2(g)4

Snowcat [4.5K]

Answer:

a

$\Delta H^o _{rxn} = -2568.9 \ kJ$

b

$H = 350 JK^{-1}$

c

$T_{max} = 32.4 ^o C$

Explanation:

From the question we are told that

The reaction of cyclobutane and oxygen is

$C_4H_8_{(g)} + 6 O_2_{(g)} \to 4 CO_2_{(g)} + 4 H_2O_{(g)}$

ΔH°f (kJ mol-1) : C4H8(g) = 27.7 ; CO2(g) = -393.5 ; H2O(g) = -241.8 ΔH° = kJ

Generally ΔH° for this reaction is mathematically represented as

$\Delta H^o _{rxn} = [[4 * \Delta H^o_f (CO_2_{(g)} ) + 4 * \Delta H^o_f(H_2O_{(g)} ] -[\Delta H^o_f (C_2H_6_{(g)} + 6 * \Delta H^o_f (O_2_{(g)}) ] ]$

=> $\Delta H^o _{rxn} = [[4 * (-393.5) + 4 * (-241.8) ] -[ 27.7 + 6 * 0]$

=> $\Delta H^o _{rxn} = -2568.9 \ kJ$

Generally the total heat capacity of 4 mol of CO2(g) and 4 mol of H2O(g), using CCO2(g) = 37.1 J K-1 mol-1 and CH2O(g) = 33.6 J K-1 mol-1. C = J K-1 is mathematically represented as

$H = [ 4 * C_{CO_2_{(g)}} + 6* C_{CH_2O_{(g)}}]$