3 years ago

4. The cloud of interstellar dust and gas that forms a star is known as

1 answer:

6 0

It's called a nebula or nebulae (plural). They are not only massive clouds of dust, hydrogen and helium gas, and plasma; they are also often “stellar nurseries” – i.e. the place where stars are born.

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If I have 9 moles of a gas at a pressure of 8.7 atm and a volume of 18 liters, what is the temperature?

alukav5142 [94]

212K
pV=nRT.
R = 0.08206

5 0

4 years ago

Consider the following reaction at equilibrium. What will happen if SO2SO2 is added to the reaction? 4FeS2FeS2(s) + 11O2O2(g) ⇌⇌

kupik [55]

Answer:

The equilibrium will change in the direction of the reactants

Explanation:

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3 years ago

How much potassium nitrate has to dissolve in water to absorb 105 kj of heat? there are 4.184 kj/kcal. also note the answer is t

joja [24]

Lattice energy of potassium nitrate (KNO3) = -163.8 kcal/mol

Hydration energy of KNO3 = -155.5 kcal/mole

Heat of solution is the amount of heat absorbed by water when 1 mole of KNO3 is dissolved in it

Heat of solution = Hydration energy - Lattice energy

= -155.5 -(-163.8) = 8.3 kcal/mol

1 kcal/mol = 4.184 kJ/mole

Therefore, 8.3 kcal/mole = 4.184 * 8.3 = 34.73 kJ/mol

Now, 34.73 kJ of heat is absorbed when 1 mole of KNO3 is dissolved

The given 105 kJ of heat would correspond to : 105/34.73 = 3.023 moles of KNO3

Molar mass of KNO3 = 101.1 g/mole

Mass of KNO3 = Molar mass * moles

= 101.1 * 3.023 = 305.63 g = 0.3056 kg

8 0

4 years ago

PLEASE HELP!!!! I NEED TO TURN IT IN TODAY!!

In-s [12.5K]

Answer:

hey mines due today did u ever get the answerrr!!

Explanation:

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3 years ago

A 36.2 g object has a heat capacity of 12.5 J/ oC. How much energy (in J) is required to raise the temperature by 5.73 oC?

iren2701 [21]

Answer:

$\Delta H=2592.825\ J$

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$ is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of object = 36.2 g

Specific heat = 12.5 J/g°C

$\Delta T=5.73\ ^0C$

So,

$\Delta H=36.2\times 12.5\times 5.73\ J=2592.825\ J$

$\Delta H=2592.825\ J$

7 0

3 years ago

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