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2 years ago

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Which statement correctly describes a scientific theory? A. It is accepted as true indefinitely. B. It is based on repeated inve

stigations. C. It is equivalent to a scientific law. D. It is accepted as false.

1 answer:

Gwar [14]2 years ago

4 0

The correct answer is B

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A partir de lo propuesto por Bohr en 1913, que postulado es correcto:

TEA [102]

Answer:

Primer postulado:

Así Bohr asumió que el átomo de hidrógeno puede existir solo en ciertos estados discretos, los cuales son denominados estados estacionarios del átomo. En el átomo no hay emisión de radiación electromagnética mientras el electrón no cambia de órbita.

Explanation:

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2 years ago

Which group of elements shares characteristics with both metals and nonmetals?

sertanlavr [38]

The answer to your question is,

Metalloids. They are a mix of elements that are both metals and non-metals in one.

-Mabel <3

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2 years ago

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What is the best statement about the data collected in Amir’s table?

Strike441 [17]

Answer: The answer is D :)

Explanation:

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3 years ago

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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

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

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

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

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0

2 years ago

Empirical formula of a compound composed of 32.1 g potassium (k) and 6.57 g oxygen (o)?

tester [92]

The empirical formula is K₂O.

The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.

The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.

So, our job is to calculate the <em>molar ratio</em> of K to O.

Step 1. Calculate the <em>moles of each element </em>

Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K

Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0

Step 2. Calculate the <em>molar ratio of each elemen</em>t

Divide each number by the smallest number of moles and round off to an integer

K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1

Step 3: Write the <em>empirical formula </em>

EF = K₂O

5 0

2 years ago