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3 years ago

What is wrong in Dalton's atomic theory?

Chemistry

1 answer:

damaskus [11]3 years ago

6 0

He decided that atoms could not be separated into smaller pieces, they can. Please mark Brainliest!!!

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Difinition of homo sapiens

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Homo sapiens is a scientific name for humans

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4 years ago

If 10.5 g copper chloride react with 12.4 g aluminum what is the limiting reactant

Brut [27]

Molar mass of copper chloride is 134.45 g/mole, so the mole of 10.5 g copper chloride is 10.5/134.45 = 0.078 mole. Molar mass of aluminum is 27 g/mole, so the mole of 12.4 g aluminium is 12.4/27 = 0.46 mole. The formula of this reaction is as follows, 2Al + 3CuCl2 ⇒ 2AlCl3 + 3Cu. Thus the molar ratio of the reactants is Al:CuCl2 = 2:3. So to react with 0.078 mole of copper chloride, will need 0.078 x 2/3 = 0.052 moles of aluminum which is less than the given amount (0.46mole). Therefore, copper chloride is the limiting reactant.

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3 years ago

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A person's temperature is 40° C. What would it be in Kelvin?

Jobisdone [24]

What do you mean.....kelvin?

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3 years ago

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Substance ΔG°f(kJ/mol) M3O4(s) −8.80 M(s) 0 O2(g) 0 Consider the decomposition of a metal oxide to its elements, where M represe

baherus [9]

Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

$K =0.02867$

Now substituting this into the equation above to obtain the equilibrium of oxygen

$0.02867 = [P_{O_2}]^{[\frac{3}{2} ]}$

multiplying through by $1 ^{\frac{2}{3} }$

$P_{O_2} = [0.02867]^{\frac{2}{3} }$

$P_{O_2} = 0.09367\ atm$

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3 years ago

Set Grouper to 70%. How do you think this level of fishing will affect the populations of the other fish in the simulated reef?

nordsb [41]

Answer:

Greatly affect.

Explanation:

This level of fishing will greatly affect the populations of the other fish in the simulated reef because overfishing disturbs the equilibrium in the marine ecosystem. One fish is a food of another fish or the big fish act as a controller which controls the population of other small fishes and in that way the ecosystem is present in equilibrium state so overfishing will greatly affect the marine ecosystem.

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3 years ago

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