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4 years ago

11

which model of an atom first introduced the idea of an nucleus? A. Dalton model, B.Thomson model, C. Rutherford model ,or D. Boh

r's model

1 answer:

frez [133]4 years ago

3 0

The answer is Rutherford's model

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Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole

trapecia [35]

Answer : The enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

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

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

The equilibrium reaction follows:

$C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)$

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

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol$

Therefore, the enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

6 0

4 years ago

Read 2 more answers

How does fabric softener affect the flammability of clothes?

Brrunno [24]

It makes your clothes smell great

6 0

4 years ago

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Using the diagram above, answer the following questions:

STatiana [176]

Answer:

See explanation for details

Explanation:

6) C is false. What is going on there is actually a transformation of mechanical energy to the chemical energy stored in carbon dioxide in the atmosphere. The burning of the fossil fuel produces mechanical energy which is stored in the chemical bonds of CO2 hence the statement is false

7) A is true. Solar energy refers to energy from the sun, this energy is trapped from Tue sun and stored in chemical bonds in glucose, and starch molecules during photosynthesis.

8) C represents the release of carbon dioxide by the burning of fossil fuels.

9) B,C,D, E

B refers to photosynthesis in which plants take in atmospheric carbon dioxide

C refers to the burning of fossil rules which releases atmospheric carbon dioxide

D refers to the dissolution of atmospheric carbon dioxide in water bodies to form carbonates in water.

E refers refers the return of organic carbon to the soil by dead organisms

10) A and C

A shows the conversion of solar energy to chemical energy

C shows the conversion of mechanical energy to chemical energy

7 0

3 years ago

At 25 °C, an aqueous solution has an equilibrium concentration of 0.00253 M for a generic cation, A2+(aq), and 0.00506 M for a g

aleksandrvk [35]

Answer:

The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M

Explanation:

Step 1: The balanced equation

AB2 ⇒ A2+ + 2B-

Step 2: Given data

Concentration of A2+ = 0.00253 M

Concentration of B- = 0.00506 M

Step 3: Calculate the equilibrium constant

Equilibrium constant Ksp of [AB2] = [A2+][B-]²

Ksp = 0.00253 * 0.00506² = 6.4777 *10^-8 M

The equilibrium constant Ksp of the generic salt AB2 = 6.4777 *10^-8 M

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4 years ago

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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

<u> $r=[A]^{1}[B]^{2}$ .............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

7 0

3 years ago