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Temka [501]
3 years ago
14

Temperature depends on the average kinetic energy of athe particles in a sample how does this help with explaining why a plasma

based light does not have a high temperature
Chemistry
2 answers:
ddd [48]3 years ago
8 0

Answer:

Most particles in the plasma are heavy and slow.

Explanation:

Hello,

In this case, one must remember that the temperature has a directly proportional relationship with the kinetic energy which is related with the energy content and motion of the molecules composing a system. In such a way, since plasma particles are heavy and slow, the kinetic energy associated with them is low which implies a low temperature as well when using them, in this manner, that low temperature helps plasma based light to be efficient with no high heating unlike traditional ways to produce light with electric energy circuits.

Best regards.

Kryger [21]3 years ago
4 0
Most particles in the plasma are heavy and slow and this statement explains why a plasma based light does not have a high temperature. The correct option among all the options that are given in the question is the third option. It is a fact that a heavy nuclei always takes more time to reach a thermal equilibrium. 

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Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (
aleksley [76]

<u>Answer:</u> The amount of heat produced by the reaction is -21.36 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

We are given:

\Delta H_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H_f_{(SO_3(g))}=-395.7kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(2\times (-395.7))]-[(2\times (-296.8))+(1\times (0))]\\\\\Delta H_{rxn}=-197.8kJ/mol

To calculate the number of moles, we use ideal gas equation, which is:

PV=nRT

where,

P = pressure of the gas = 1.00 bar

V = Volume of the gas = 2.67 L

n = number of moles of gas = ?

R = Gas constant = 0.0831\text{ L. bar }mol^{-1}K^{-1}

T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

To calculate the heat released of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.108 moles

\Delta H_{rxn} = enthalpy change of the reaction = -197.8 kJ/mol

Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

3 0
3 years ago
Write the chemical symbol for the isotope of cobalt with 37 neutrons
oksano4ka [1.4K]

Answer:

64Co27

Explanation:

4 0
3 years ago
A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be
Ymorist [56]

Answer:

a)    [Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)    1.273 × 10⁻¹⁶

c)    2.629×10⁻¹⁹ M Thus; the value for  [Ag+ ]dilute will be too low

Explanation:

In an Ag | Ag+ concentration cell ,

The  anode reaction can be written as :

Ag ----> Ag+(dilute) + e-    &:

The  cathode reaction can be written as:

Ag+(concentrated) + e- ----> Ag

The  Overall Reaction : is

Ag+(concentrated) -----> Ag+(dilute)

However, the Standard Reduction potential of cell = E°cell = 0

( since both cathode and anode have same Ag+║Ag )

Also , given that the theoretical slope is - 0.0591 V

Therefore; the reduction potential of cell ; i.e

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963  

[Ag+]dilute = \mathbf{10^{-14.1963} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 6.363  × 10⁻¹⁶ M  

b)

AgI ----> Ag + (dilute) + I⁻

So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]  

= 6.363 × 10⁻¹⁶ M × 0.20 M  

= 1.273 × 10⁻¹⁶

c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :

Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )

1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )  

log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804  

[Ag+]dilute = \mathbf{10^{-17.5804} } × 1.0 × 10⁻¹ M

[Ag+]dilute = 2.629×10⁻¹⁹ M

Thus, the value for  [Ag+ ]dilute will be too low

5 0
3 years ago
some inkjet printers produce picoliter-sized drops. how many water molecules are there in one picoliter of water? the density of
GrogVix [38]

3.37 x 10¹⁰ molecules

Explanation:

Given parameters:

Volume of water = 1pL = 1 x 10⁻¹²L

Density of water = 1.00g/mL = 1000g/L

Unknown:

Number of water molecules = ?

Solution:

To solve this problem, we first find the mass of the water molecule in the inkjet.

       Mass of water = density of water x volume of water

Then, the number of molecules can be determined using the expression below:

        number of moles = \frac{mass of water}{molar mass of water}

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Solving:

Mass of water = 1 x 10⁻¹² x 1000 = 1 x 10⁻⁹g

Number of moles:

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Number of moles = \frac{1 x 10^{-12} }{18} = 5.6 x 10⁻¹⁴moles

Number of molecules =  5.6 x 10⁻¹⁴   x   6.02 x 10²³ = 33.7 x 10⁹

                                     = 3.37 x 10¹⁰ molecules

Learn more:

Number of molecules brainly.com/question/4597791

#learnwithBrainly

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3 years ago
What role in a factory does the cell membrane play
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The plasma membrane
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3 years ago
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