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Agata [3.3K]
3 years ago
6

How would multiplication affect a conversion factor

Chemistry
1 answer:
hichkok12 [17]3 years ago
3 0

Can u be a little more specific so that I can help u

I will edit when u specify a little more.

Have a great day.

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Lelu [443]

Answer: is is true

Explanation:

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2 years ago
I NEED HELP WITH NUMBER 3
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The C stands for Specific Heat
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The enthalpy of vaporization of Bromine is 15.4 kJ/mol. What is the energy change when 80.2 g of Br2 condenses to a liquid at 59
Aleksandr-060686 [28]

The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.

<h3>What is Enthalpy of Vaporization ?</h3>

The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.

<h3>How to find the energy change from enthalpy of vaporization ?</h3>

To calculate the energy use this expression:

Q = n \Delta H_{\text{vapo.}

where,

Q = Energy change

n = number of moles

\Delta H_{\text{Vapo.}} = Molar enthalpy of vaporization

Now find the number of moles

Number of moles (n) = \frac{\text{Given Mass}}{\text{Molar mass}}

                                   = \frac{80.2\ g }{159.8\ g/mol}

                                   = 0.5 mol

Now put the values in above formula we get

Q = - n \Delta H_{\text{vapo.}         [Negative sign is used because Br₂ condensed here]

   = - (0.5 mol × 15.4 kJ/mol)

   = - 7.7 kJ

Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.

Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849

#SPJ1

8 0
2 years ago
PLEASE HELP
Dovator [93]

Answer:

B

Explanation:

All elements on both sides of the equation have the same amount of moles

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You see the color orange in the neon sign above because electrified neon atoms produce a single orange colored wavelength of lig
Bezzdna [24]

Answer: As a result, each excited electron in an atom emits a photon of a specific wavelength. To put it another way, each excited noble gas emits a distinct hue of light. This is a reddish-orange neon light.

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