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3 years ago

Which is an example of a highly unstable isotope that is often used in fission reactions?

2 answers:

4 0

The right answer to this question is U-235. It is an isotope of Uranium and it is an unsteady heavy metal used in fission reactions because it can run long chains of reactions. I hope this helps.

Marta_Voda [28]3 years ago

4 0

Answer:

C. U-235

Explanation:

A fission reaction is one in which an unstable radioisotope breaks down into smaller nuclei when bombarded with energetic particles.

The highly unstable isotope that is often used in fission reactions is U-235. When bombarded with neutrons that atom breaks down to form krypton and barium along with the release of a large amount of energy. The reaction is given as:

₉₂U²³⁵ + ₀n¹ → ₃₆Kr⁸⁹ + ₅₆Ba¹⁴⁴ + 3₀n¹ + 210 Mev energy

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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta

shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]$

Putting the values we get :

$\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]$

$\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]$

$\Delta H=-5314.8kJ$

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = $\frac{5314.8}{2}\times 1=2657.4kJ$

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0

2 years ago

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$

6 0

3 years ago

15.

lorasvet [3.4K]

C.

Because the spinal cord functions primarily in the transmission of neural signals between the brain and the rest of the body

5 0

2 years ago

If i have 8 L of 0.25 M solution, how many miles of MgCl2 are present

qaws [65]

Answer:

2 moles

Explanation:

The following were obtained from the question:

Molarity = 0.25 M

Volume = 8L

Mole =?

Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:

Molarity = mole of solute/Volume of solution.

With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:

Molarity = mole of solute/Volume of solution.

0.25 = mole of MgCl2 /8

Cross multiply to express in linear form

Mole of MgCl2 = 0.25 x 8

Mole of MgCl2 = 2 moles

Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution

4 0

2 years ago

Sample A: 300 mL of 1M sodium chloride

yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

volume of sample B, V = 145 mL = 0.145 L

Molarity of sample B, C = 1.5 M

molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol

Molarity is given as;

$C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol$

The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g

The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g

The density of sample A $= \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L$

The density of sample B $= \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L$

Therefore, sample B contains the larger density

5 0

2 years ago