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otez555 [7]
4 years ago
11

I WILL GIVE BRAINLIEST!

Chemistry
2 answers:
Triss [41]4 years ago
8 0

An electron.

Now, Brainliest?

anastassius [24]4 years ago
6 0

The correct answer is A : Electron.

An Electron is accepted by an atom when it reacts with other atoms.

Hope I Helped!

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An ongoing series of fission reactions is called a (n) chain reaction.   
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What is the molecule polarity of SiCl4????
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Silicon Tetrachloride:Explanation:

Silicon tetrachloride has the formula SiCl4. It is a covalent compound, which means that electrons are shared between the silicon atom and the chlorine atoms. The bond angles between the chlorine atoms and the silicon depend on the molecule's shape.

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1. Who is the father of atomic theory? 2. Who discovered the electron? 3. Who expressed particles by wave equations? 4. Who rese
krek1111 [17]
1. Who is the father of atomic theory?
Dalton

2. Who discovered the electron?
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3. Who expressed particles by wave equations?
<span>Schrödinger
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<span>5. Who discovered the "open spaces" model?
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6. Who applied quantum theory to atoms? 
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6 0
4 years ago
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Why is radioactive dating important when approximating the age of earth?
Furkat [3]

Answer:

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4 years ago
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Ethanol has a heat of vaporization of 38.56kj/mol and a normal boiling point of 78.4 ∘c.
Elanso [62]

This is an incomplete question, here is a complete question.

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 14 °C?

Answer : The vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 14.0^oC = ?

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm

T_1 = temperature of ethanol = 14.0^oC=273+14.0=287K

T_2 = normal boiling point of ethanol = 78.4^oC=273+78.4=351.4K

\Delta H_{vap} = heat of vaporization = 38.56 kJ/mole = 38560 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})

P_1=5.174\times 10^{-2}atm

Hence, the vapor pressure of ethanol at 14.0^oC is 5.174\times 10^{-2}atm

4 0
3 years ago
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