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3 years ago

Which applies to fision

Chemistry

2 answers:

pantera1 [17]3 years ago

7 0

a) involves the splitting of nuclei

b) releases large amounts of energy

D) releases radiation as a waste product

e) occurs in nuclear power plants and is used to generate electricity

lara [203]3 years ago

6 0

Wasssssssupppp with you

You might be interested in

In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.00 degrees C. After

igomit [66]

Answer:

+26.6kJ/mol

Explanation:

The enthalpy of dissolution of NH₄NO₃ is:

NH₄NO₃(aq) + ΔH → NH₄⁺ + NO₃⁻

Where ΔH is the heat of reaction that is absorbed per mole of NH₄NO₃,

The moles that reacts in 1.60g are (Molar mass NH₄NO₃:80g/mol):

1.60g * * (1mol / 80g) = 0.02 moles reacts

To find the heat released in the coffee cup calorimeter, we must use the equation:

Q = m×ΔT×C

Where Q is heat released,

m is mass of the solution

ΔT is change in temperature (Final temperature - Initial temperature)

C is specific heat of the solution (4.18J/g°C)

Mass of the solution is:

1.60g + 75g = 76.60g

Change in temperature is:

25.00°C - 23.34°C = 1.66°C

Replacing:

Q = m×ΔT×C

Q = 76.60g×1.66°C×4.18J/g°C

Q = 531.5J

This is the heat released per 0.02mol. The heat released per mole (Enthalpy change for the dissolution of NH₄NO₃) is:

531.5J / 0.02mol = 26576J/ mol =

+26.6kJ/mol

+ because the heat is absorbed, the reaction is endothermic-

7 0

3 years ago

Which Two of the following are pure substances*

Snowcat [4.5K]

Answer:

I think A and B

Explanation:

5 0

3 years ago

Read 2 more answers

A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure

Alex

Answer:

B.3/5p

Explanation:

For this question, we have to remember "Dalton's Law of Partial Pressures". This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.

Additionally, we have a proportional relationship between moles and pressure. In other words, more moles indicate more pressure and vice-versa.

$P_i=P_t_o_t_a_l*X_i$

Where:

$P_i$ =Partial pressure

$P_t_o_t_a_l$ =Total pressure

$X_i$ =mole fraction

With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:

moles of hydrogen gas

The molar mass of hydrogen gas ( $H_2$ ) is 2 g/mol, so:

$6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2$

moles of oxygen gas

The molar mass of oxygen gas ( $O_2$ ) is 32 g/mol, so:

$64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2$

Now, total moles are:

Total moles = 2 + 3 = 5

With this value, we can write the partial pressure expression for each gas:

$P_H_2=\frac{3}{5}*P_t_o_t_a_l$

$P_O_2=\frac{2}{5}*P_t_o_t_a_l$

So, the answer would be 3/5P.

I hope it helps!

5 0

3 years ago

1. The heat of fusion for the ice-water phase transition is 335 kJ/kg at 0°C and 1 bar. The density of water is 1000 kg/m3 at th

vodomira [7]

Answer:

Expression for the change of melting temperature with pressure..> T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa), Freezing Point = 0°C

Explanation:

Derivation from state postulate

Using the state postulate, take the specific entropy, , for a homogeneous substance to be a function of specific volume and temperature .

ds = (partial s/partial v)(t) dv + (partial s/partial T)(v) dT

During a phase change, the temperature is constant, so

ds = (partial s/partial v)(T) dv

Using the appropriate Maxwell relation gives

ds = (partial P/partial T)(v) dv

s(β) – s(aplαha) = dP/dT (v(β) – v(α))

dP/dT = s(β) – s(α)/v(β) – v(α) = Δs/Δv

Here Δs and Δv are respectively the change in specific entropy and specific volume from the initial phase α to the final phase β.

For a closed system undergoing an internally reversible process, the first law is

du = δq – δw = Tds - Pdv

Using the definition of specific enthalpy, h and the fact that the temperature and pressure are constant, we have

du + Pdv = dh Tds,

ds = dh/T,

Δs = Δh/T = L/T

After substitution of this result into the derivative of the pressure, one finds

dp/dT = L/TΔv

This last equation is the Clapeyron equation.

(dP/dT) = dH/TdV => dP/dlnT = dH/dV

=> dP/dlnT = dH/dV = [H(liquid) - H(solid)]/[V(liquid) - V(solid)]

= [335,000 J/kg]/[1000⁻¹ - 915⁻¹ m³/kg]

= -3.61x10⁹ J/m³ = -3.61x10⁹ Pa

=> P₂ = P₁ - 3.61x10⁹ ln(T₂/T₁) Pa

T₂ = T₁exp(-(P₂-P₁)/(3.61x10⁹ Pa)

b) if the pressure in Denver is 84.6 kPa:

T₂(freezing) = 273.15exp[-(84,600-100,000)/(3.61x10⁹)]

≅ 273.15 = 0°C T₁(freezing) essentially no change

5 0

3 years ago

Which two changes would make this reaction reactant-favored?

Vera_Pavlovna [14]

Two changes would make this reaction reactant-favored

C. Increasing the temperature

D. Reducing the pressure

<h3>Further explanation</h3>

Given

Reaction

2H₂ + O₂ ⇒ 2H₂0 + energy

Required

Two changes would make this reaction reactant-favored

Solution

The formation of H₂O is an exothermic reaction (releases heat)

If the system temperature is raised, then the equilibrium reaction will reduce the temperature by shifting the reaction in the direction that requires heat (endotherms). Conversely, if the temperature is lowered, then the equilibrium shifts to a reaction that releases heat (exothermic)

While on the change in pressure, then the addition of pressure, the reaction will shift towards a smaller reaction coefficient

in the above reaction: the number of coefficients on the left is 3 (2 + 1) while the right is 2

As the temperature rises, the equilibrium will shift towards the endothermic reaction, so the reaction shifts to the left towards H₂ + O₂( reactant-favored)

And reducing the pressure, then the reaction shifts to the left H₂ + O₂( reactant-favored)⇒the number of coefficients is greater

4 0

2 years ago