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Mama L [17]
2 years ago
15

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a

neutron:proton ratio of 1.05
Chemistry
1 answer:
kvv77 [185]2 years ago
5 0

The question is incomplete, the complete question is;

Which of the following is most likely a heavier stable nucleus? (select all that apply) Select all that apply: A nucleus with a neutron:proton ratio of 1.05 A nucleus with a A nucleus with a neutron:proton ratio of 1.49 The nucleus of Sb-123 A nucleus with a mass of 187 and an atomic number of 75

Answer:

A nucleus with a A nucleus with a neutron:proton ratio of 1.49

A nucleus with a mass of 187 and an atomic number of 75

Explanation:

The stability of a nucleus depends on the number of neutrons and protons present in the nucleus. For many low atomic number elements, the number of protons and number of neutrons are equal. This implies that the neutron/proton ratio = 1

Elements with higher atomic number tend to be more stable if they have a slight excess of neutrons as this reduces the repulsion between protons.

Generally, the belt of stability for chemical elements lie between and N/P ratio of 1 to an N/P ratio of 1.5.

Two options selected have an N/P ratio of 1.49 hence they are heavy stable elements.

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Oxygen at 150 k and 41.64 atm has a tabulated specific volume of 4.684 cm3/g and a specific internal energy of 1706 j/mol. calcu
docker41 [41]

The total energy in a system due to the temperature and pressure per unit mass in that system is known as specific enthalpy. It is used in thermodynamic equations when one desires to know the energy for a given single unit mass of a component.  

Specific enthalpy is calculated by the equation:  

H = U + PV

in the given case, Specific volume = 4.684 cm³/g = 149.888 cm³/g moles = 149.888 × 10⁻³ J/g moles

Specific internal energy (U) is 1706 J/mol and pressure is 41.64.  

H = 1706 + 41.64 × 149.888 × 10⁻³ × 101.3 joules  

= 2428 joules / mole


5 0
2 years ago
How much heat is released or absorbed (in kJ) by the system in the reaction of 17.7 g of SF6 with 23.7 g of H2O?
Jlenok [28]

Answer:

+279.744 kJ.

Below is an attachment containing the solution to question.

5 0
2 years ago
Bobby's mom sets a small pot of water on the stove and lights the burner. Ten minutes later, Bobby notices small bubbles and see
dolphi86 [110]
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7 0
2 years ago
Read 2 more answers
A gas has a volume of 22.4 L at 0.853 atm. What pressure is needed to change the volume to 24.0L?
Scrat [10]

The pressure needed to change the volume to 24 L is 0.796 atm.

Explanation:

It is known by Boyle's law that the pressure experienced by the gas molecules will be inversely proportional to the volume occupied by the molecules.

P =\frac{1}{V}

So as the initial volume is said to be 22.4 L, consider it as V₁ = 22.4 L. Then the initial pressure is said to be 0.853 atm, so P₁ = 0.853 atm. So we have to determine the new pressure P₂ when the volume is changed to V₂ = 24 L.  As there is increase in the volume, the pressure should be decreased due to Boyle's law. Thus, as per Boyle's law, the two pressures and their volumes can be related as

P_{1} V_{1} =  P_{2} V_{2}

0.853*22.4 = P_{2}*24\\\\P_{2} = \frac{0.853*22.4}{24} = 0.796 atm

Thus, the pressure gets decreased to 0.796 atm on increase in the volume to 24 L.

So the pressure needed to change the volume to 24 L is 0.796 atm.

6 0
2 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
2 years ago
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