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Veronika [31]
3 years ago
13

Help

Chemistry
1 answer:
Rudik [331]3 years ago
7 0

Answer: One change of state happens when you add energy to the substance. This change of state is called melting. By adding energy to the molecules in a solid the molecules begin to move quicker and can break away from the other molecules.

Explanation: I hoped that helped!

You might be interested in
he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

8 0
3 years ago
Molecule contains carbon, hydrogen and sulfur atoms. When a sample of 0.535g of this compound is burnt in oxygen, 1.119 g of CO2
OLga [1]

Answer:

The empirical formula is, C4H4S

Explanation:

Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles

Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g

Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles

Mass of hydrogen = 0.025 moles × 1 = 0.025 g

Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles

Mass of sulphur= 0.0064 moles ×32 = 0.2 g

Now we obtain the mole ratios by dividing through by the lowest ratio.

C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles

C4H4S

4 0
3 years ago
How many Mg atoms are presented in 3.00 moles of MgCl2
ella [17]

Answer:

1.81 x 10²⁴ atoms

Explanation:

To find the number of atoms in the given number of moles, we need to understand that every substance contains the Avogadro's number of particles.

More appropriately, a mole of any substance will contain the Avogadro's number of particles which is 6.02 x 10²³ atoms

 So;

           If 1 mole of a substance  = 6.02 x 10²³ atoms;

               3 mole of MgCl₂ will contain 3 x 6.02 x 10²³ = 1.81 x 10²⁴ atoms

3 0
3 years ago
Titanium has five common isotopes: 46Ti (8.0%), 47Ti (7.8%), 48Ti (73.4%), 49Ti (5.5%), 50Ti (5.3%). What is the average atomic
NNADVOKAT [17]

Hey there!:

Isotopes :                          abundance :

46 Ti                                       8.0%

47 Ti                                        7.8 %

48 Ti                                      73.4 %

49 Ti                                       5.5 %

50 Ti                                         5.3 %

Weighted average =   ∑ Wa * % / 100

Therefore:

( 46 * 8.0) + (47 * 7.8 ) + (48 * 73.4 ) + ( 49 * 5.5 ) + ( 50*5.3 ) / 100 =

4792.3 / 100

= 47.923 a.m.u


       Hope that helps!

7 0
3 years ago
What is the energy of an electromagnetic wave with a frequency of 8•10^12 Hz?
aniked [119]

Hello!

Find the Energy of the Photon by Planck's Equation, given:

E (photon energy) =? (in Joule)

h (Planck's constant) = 6.626*10^{-34}\:J * s

f (radiation frequency) = 8*10^{12}\:Hz

Therefore, we have:

E = h*f

E = 6.626*10^{-34}*8*10^{12}

E = 53.008*10^{-34+12}

E = 53.008*10^{-22}

\boxed{\boxed{E = 5.3008*10^{-21}\:Joule}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

8 0
3 years ago
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