A real gas behaves most like an ideal gas at?

What is the most likely reason a computer model is used for a black hole?

Rashid [163]

Answer:

Black holes are merely the most exotic example of the general principle that ... quantum computer stores bits on protons and uses magnetic fields to flip them. ... Powered by Standard Model software, the universe computes

Explanation:because they are mostly exotic

6 0

3 years ago

?cual es el tiempo de<br>onda con 6<br>oscilaciones.?

Setler79 [48]

Answer:

Hey hi

Explanation:

Can you pls tell me which language is this.... Pls really sorry... I wanna help you

7 0

3 years ago

A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200

34kurt

Answer:

81 °C

Explanation:

This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:

q = heat added/released by a sample

m = mass of sample

c=specific heat of sample

ΔT = change in temperature

from here we can rearrange the equation to state:

q/(mc) = ΔT

1200J/((20.0g)(4.2J/g•°C)) = ΔT

14°C = ΔT

If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.

4 0

3 years ago

Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) given +142.7 kJ/mol for the standard enthalpy of

bulgar [2K]

Answer:

The standard reaction enthalpy for the given reaction is 235.15 kJ/mol.

Explanation:

$O_2(g) \rightarrow \frac{2}{3}O_3(g),\Delta H^o_{1}=142.7 kJ/mol$ ..[1]

$O_2(g) \rightarrow 2 O(g),\Delta H^o_{2}=498.4 kJ/mol$ ..[2]

$NO(g) + O_3(g)\rightarrow NO_2(g) + O_2(g) ,\Delta H^o_{3} = -200 kJ/mol$ ..[3]

$NO_2(g)\rightarrow NO(g) + O(g),\Delta H^o_{4}=?$ ..[4]

Using Hess's law:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

2 × [4] = [2]- (3 ) × [1] - (2) × [3]

$2\times \Delta H^o_{4}=\Delta H^o_{2} -3\times \Delta H^o_{1}-2\times \Delta H^o_{3}$