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gavmur [86]
3 years ago
14

1. In general, the nucleus of a small atom is stable. Therefore, over very short distances, such as those in a small nucleus,

Chemistry
2 answers:
Dafna1 [17]3 years ago
6 0
1. In general, the nucleus of a small atom is stable. Therefore, over very short distances, such as those in a small nucleus,

a. the strong nuclear force is much greater than the electric force.


9. Uranium-238 undergoes alpha decay. Therefore, uranium-238 will


b. change into a different element altogether.

Hope these answer the questions. Have a nice day.

postnew [5]3 years ago
5 0
<h3>1. <u>Answer;</u></h3>

a. the strong nuclear force is much greater than the electric force.

<h3><u>Explanation</u>;</h3>
  • <em><u>For an atom to be stable it means it has enough amount of binding energy to hold its nucleus together permanently. </u></em>
  • Therefore, <em><u>an unstable atom lacks enough amount of binding energy to hold its nucleus permanently and thus undergoes decay to achieve stability. Unstable atoms are therefore referred to being radioactive.</u></em>
  • Small atoms are stable; <u>this is because they have equal number of protons and neutrons and thus the protons and neutrons fill up energy levels while maximizing the strong force binding the nucleus together. </u>

<h3>9.<u> Answer;</u></h3>

b. change into a different element altogether.

Uranium-238 undergoes alpha decay. Therefore, uranium-238 will  <em><u>change into a different element altogether</u></em>.

<h3><u> Explanation;</u></h3>
  • Unstable atoms undergo radioactive decay in order to achieve stability of their nucleus.
  • <em><u>Uranium-238 is an example of such atom, which may undergo decay to achieve stability.</u></em>
  • <em><u>Alpha decay is one of the types of decays,</u></em> others being beta decay and gamma decay. <em><u>In alpha decay the radioactive isotope undergoes decay such that its mass number is decreased by four and its atomic number is decreased by two.</u></em>
  • Therefore, <em><u>Uranium-238 undergoes alpha decay to form a different element whose mass number is 234 and atomic number is 90, known as thorium-234. </u></em>
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A substance has a mass of 23.7g and a volume of 7.9ml. The density of the object is
Nataly [62]

Answer:

3 grams per millliter

Explanation:

\rho=\dfrac{m}{V} \\\\\\\rho=\dfrac{23.7g}{7.9mL}=3g/mL

Hope this helps!

4 0
3 years ago
A 10.00 mL sample of a 1.07 M solution of potassium hydrogen phthalate (KHP, formula mass = 204.22 g/mol) is diluted to 250.0 mL
ser-zykov [4K]

Answer:

[KHP] = 0.0428M

Explanation:

2 methods to calculate concentration after dilution

1. Use dilution equation

Molarity of concentrate (M₁) x Volume of Concentrate (V₁)

           =    Molarity of dilute (M₂) x Volume of dilute (V₂)

M₁ x V₁ = M₂ x V₂ => M₂ = M₁ x V₁ / V₂ = (1.07M)(10ml)/(250ml) = 0.0428M

2. Concentration Equation

moles KHPh = Molarity (M)  x  Volume (V) = 1.07M x 0.010L =0.0107 moles KHP

Concentration KHP = moles solute / volume of solution in Liters

= 0.0107 moles KHP / 0.25L = 0.0428M

8 0
3 years ago
Liquid water at 325 K and 8000 kPa flows into a boiler at a rate of 10 kg⋅s− 1 and is vaporized, producing saturated vapor at 80
masha68 [24]

Answer:

0.4058

Explanation:

From the steam table

At 325 K, saturated liquid

Enthalpy, Hf = 217 kJ/kg

Entropy Sf = 0.7274 kJ/kg-K

Specific volume Vf = 1.013 cm3/gm

Saturated pressure Psat = 12.87 kPa

For compressed liquid

P1 = 8000 kPa

Temperature T = 325 K

Thermal expansion coefficient B = 460 x 10^{-6} K^{-1}

Enthalpy at initial conditions

H1 = Hf + Vf (1-BT)(P1 - P_sat)

= 217 + 1.013*10^-3 (1 - 460*10^{-6})(8000 - 12.87)

= 223.881 kJ/kg

Entropy at initial conditions

S1 = Sf - BVf (P1 - Psat)

= 0.7274 - 460×10^{-6}*1.013*10^-3 (8000 - 12.87)

= 0.724 kJ/kg-K

At 8000 kPa, saturated vapor

H2 = 2759.9 kJ/kg

S2 = 5.7471 kJ/kg-K

T = 300 K

Heat added Q = H_2 - H_1

= 2759.9 - 223.881

= 2536 kJ/kg

Maximum work

W = (H1 - H2) - T (S1 - S2)

= 223.881 - 2759.9 - 300(0.724 - 5.7471)

= - 1029 kJ/kg

Fraction of heat added = W/Q

= 1029/2536

= 0.4058

3 0
3 years ago
What are the structures in which stars are grouped together?
disa [49]

Answer:

constellations

Explanation:

4 0
3 years ago
The reaction of NO(g) + Cl2(g)⇌NOCl(g) + Cl(g) is found to have an equilibrium constant of 1.1×108 at a particular temperature.
irakobra [83]

Answer:

Option (2) At equilibrium, there is a much higher concentration of products than reactants.

Explanation:

The equilibrium constant for a reaction is simply the ratio of the concentration products raised to their mole ratio divided by the concentration of the reactants raised to their mole ratio.

If the equilibrium constant is close to 1 or 1, it means the concentration of the reactants and products are almost the same. But if the equilibrium constant is large as in the case of the question given above, it means that at equilibrium, the concentration of the products are higher than that of the reactants.

6 0
3 years ago
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