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Readme [11.4K]
2 years ago
13

In the PhET simulation, make sure that the checkbox Stable/Unstable in the bottom right is checked. Using the PhET simulation as

a reference, determine whether the following combinations of protons and neutrons form stable or unstable nuclei by adding protons and neutrons to the nucleus. Unstable nuclei will shake. If the Stable/Unstable checkbox in the lower right is also checked, the nuclei will be labeled Stable or Unstable. Sort the following combinations of protons and neutrons according to whether or not they form stable or unstable nuclei.
a. 4 protons and 5 neutrons
b. 7 protons and 7 neutrons
c. 2 protons and 3 neutrons
d. 3 protons and 0 neutrons
e. 6 protons and 5 neutrons
f. 9 protons and 9 neutrons
g. 8 protons and 7 neutrons
h. 1 proton and 0 neutrons
Chemistry
1 answer:
lord [1]2 years ago
6 0

Answer:

kindly check the EXPLANATION SECTION

Explanation:

In order to be able to answer this question one has to consider the neutron proton ratio. Considering this ratio will allow us to determine the stability of a nuclei. The most important rule that helps us in determination of stability is that when the Neutron- Proton ratio  of any nuclei ranges from  to 1 to 1.5, then we say the nuclei is STABLE.

Also, we need to understand that when the  Neutron- Proton ratio is LESS THAN 1 or GREATER THYAN 1.5, then we say the nuclei is UNSTABLE.

So, let us check which is stable and which is unstable:

a. 4 protons and 5 neutrons =  Neutron- proton ratio = N/P = 5/4= stable.

b. 7 protons and 7 neutrons  =  Neutron- proton ratio = N/P = 7/7= 1 = stable.

c. 2 protons and 3 neutrons  =  Neutron- proton ratio = N/P = 3/5 =0.6 =unstable.

d. 3 protons and 0 neutrons  =  Neutron- proton ratio = N/P = 0/3= 0= unstable.

e. 6 protons and 5 neutrons  =  Neutron- proton ratio = N/P = 5/6= 0.83 = unstable.

f. 9 protons and 9 neutrons  =  Neutron- proton ratio = N/P = 9/9 = 1 = stable.

g. 8 protons and 7 neutrons  =  Neutron- proton ratio = N/P =  7/8 =0.875 = unstable.

h. 1 proton and 0 neutrons =  Neutron- proton ratio = N/P = 0/1 =0 = unstable

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5.00 g of hydrogen gas and 50.0g of oxygen gas are introduced into an otherwise empty 9.00L steel cylinder, and the hydrogen is
GenaCL600 [577]
1) Balanced chemical reaction:

2H2 + O2 -> 2H20

Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O

2) Reactant quantities converted to moles

H2: 5.00 g / 2 g/mol = 2.5 mol

O2: 50.0 g / 32 g/mol = 1.5625 mol

Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).

3) Products

H2 totally consumed -> 0 mol at the end

O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end

H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.

Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol

4) Pressure

Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K

p = nRT/V  = 7.9 atm
3 0
2 years ago
Read 2 more answers
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

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2 years ago
An equimolar mixture of N2(g) and Ar(g) is kept inside a rigid container at a constant temperature of 300 K. The initial partial
andrey2020 [161]

Answer:

The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm

Explanation:

Using the ideal gas law

PV=nRT

if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore

Inicial state ) P₁V=n₁RT

Final state )  P₂V=n₂RT

dividing both equations

P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁

now we have to determine P₁ and n₂ /n₁.

For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)

p ar₁ = P₁ * x ar₁  →  P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm

n₁ = n ar₁ + n N₁ =  n ar₁ + n ar₁ = 2 n ar₁

n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁

n₂ /n₁ = 3/2

therefore

P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2  = 2.25 atm

P₂= 2.25 atm

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2 years ago
A 6.00 g sample of calcium sulfide is found to contain 3.33 g of calcium. what is the percent by mass of sulfur in the compound?
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What is the concentration of an alcl3 solution if 150. ml of the solution contains 550. mg of cl- ion?
valina [46]

The concentration of AlCl3 solution if 150 ml of the solution contains 550 mg of cl- ion is 0.0344 M


calculation


concentration = moles /volume in liters


volume in liters = 150 /1000= 0.15 L


number of moles calculation

write the equation for dissociation of Al2Cl3

that is AlCl3 ⇔ Al^3+ + 3 Cl ^-


find the moles of Cl^- formed

moles =mass/molar mass

mass in grams= 550/ 1000 =0.55 grams

molar mass of Cl^- =35.5 g/mol


moles is therefore= 0.55/35.5 =0.0155 moles


by use of mole ration betweem AlCl3 to Cl^- which is 1:3 the moles of AlCl3 is =0.0155 x 1/3= 5.167 x10^-3 moles



concentration of AlCl3 is therefore= 5.167 x10^-3/ 0.15 =0.0344 M

6 0
2 years ago
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