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4 years ago

What determines the life cycle of a star

1 answer:

4 0

Hi There! :)

What determines the life cycle of a star?

A star's life cycle is determined by its mass.

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determine the percent yield for carbon dioxide if 4.50 moles of propane yielded 7.64 moles of carbon dioxide

ira [324]

Answer:

Percent yield = 57%

Explanation:

Given data:

Number of moles of propane = 4.50 mol

Number of moles of carbon dioxide = 7.64 mol

Percent yield = ?

Solution:

Chemical equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Now we will compare the moles of propane and carbon dioxide.

C₃H₈ : CO₂

1 ; 3

4.50 : 3×4.50 = 13.5 mol

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 7.64 mol / 13.5 mol × 100

Percent yield = 0.57× 100

Percent yield = 57%

3 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago

Solve the enthalpy of formation NH3(g) -46.11

lora16 [44]

I think 43.12 I’m not that good with math

3 0

4 years ago

If a 2,000-kilogram car accelerates at a rate of 3 meters

Nostrana [21]

Explanation:

Newton's second law of motion states that the acceleration of an object is dependant on the amount of force applied and the mass of the object itself, which is illustrated by the equation

$F \ = \ ma$ ,

where $F$ is the force applied, $m$ is the mass of the object and $a$ is the acceleration.

Therefore,

$F \ = \ (2000 \ \text{kg})(3 \ \text{m\,s}^{-2}) \\ \\ F \ = \ 6000 \ \text{N}$

Hence, 6000 N of force is applied by the engine to the car.

5 0

3 years ago

3NO2− + 8H+ + Cr2O72− → 3NO3− +2Cr3+ + 4H2O

Reptile [31]

According to this reaction:
3NO2^- + 8 H^+ + Cr2O7^-2 → 3NO3^- + 2Cr^+3 + 4 H2O
(1) 3 NO2^- → 3NO3^-, The oxidation state of nitrogen converted from +3 to +5
i.e oxidation process, ∴ NO2^- is reducing agent.
(2) Cr2O7^-2 → 2Cr^+3, The oxidation state of Cr converted from +6 to +3
i.e reduction process ∴ Cr2O7^-2 is an Oxidizing agent.
(3) H^+ → H2O, the oxidation state of Hydrogen is still +1, i.e No change (neither)

6 0

4 years ago