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4 years ago

How was Rutherford’s gold foil experiment inconsistent with the plum pudding model of the atom?

1 answer:

maks197457 [2]4 years ago

5 0

According to the plum pudding model of the atom proposed by J.J. Thomson, the atom was considered to be a positive charge sphere in which electrons (negatively charged) are embedded in order to balance the positive charge.

According to Rutherford’s gold foil experiment, when he bombarded $\alpha$ particles to the gold foil he observed that most of the particles went through, some of them got scattered in various directions, and few are deflected back to the source. Hence, this experiment shows that the plum pudding model was incorrect as; if there was the symmetrical distribution of the charges then all the $\alpha$ particles will pass through without any deflection.

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A solution has the 23.22% mass/volume of a particular solvent what is the mass of solute dissolved in 2.5 liter of solvent?

valina [46]

23.22%:

2.5 L in mL : 2.5 * 1000 = 2500 mL

23.22 g ----------- 100 mL
? g ------------- 2500 mL

m = 2500 * 23.22 / 100

m = 58050 / 100

m = 580,5 g

3 0

3 years ago

Write the net ionic equ.

Artyom0805 [142]

Answer:

a) AgNO3 + KI → AgI + KNO3

b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O

c) 2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3

d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Explanation:

a) AgNO3 + KI → Ag+ + NO3- + K+ + I-

Ag+ + NO3- + K+ + I- → AgI + KNO3

AgNO3 + KI → AgI + KNO3

b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-

Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O

Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O

c) 2Na3PO4 + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-

6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- → Ni3(PO4)2 + 6NaNO3

2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3

d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-

2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

7 0

3 years ago

How much heat is released when 47.50 g of CH4 (g) is burned in excess oxygen gas to produce carbon dioxide and water?

Elina [12.6K]

Answer:

= 2113.44 kJ

Explanation:

When 1 mole of CH4 (g) burns in excess oxygen -714.0 kJ of heat is released.

For 47.50 g;

molar mass of CH4 = 16.042 g/mol

Number of moles of CH4

= 47.5 g /16.042 g/mol

= 2.96 moles

Therefore;

1 mole = -714.0 kJ

Heat change for 2.96 moles

= 2.96 moles × -714.0 kJ

= 2113.44 kJ

8 0

3 years ago

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound Liquid molar volume (cm³/mol)

Ethanol (1) 58.68

1 - propanol (2) 75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = $\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )$ ------------ equation (1)

where:

$a_{12}/R$ = Wilson parameter = ???

$V_2$ = liquid molar volume of component 2 = 75.14 cm³/mol

$V_1$ = liquid molar volume of component 1 = 58.68 cm³/mol

T = temperature = 25° C = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

$0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$In (0.547) = \ (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R= 0.60 * 298.15 \ K$

$-a_{12}/R= - 178.89 \ K$

$a_{12}/R= 178.89 \ K$

To calculate the temperature dependent parameters of the Wilson equation ∧₂₁

∧₂₁ = $\frac{V_1}{V_2} \ exp \ (\frac{-a_{12}/R}{T} )$ ---------- equation (2)

$1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1.1 = 0.7809 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$\frac{1.1}{0.7809}= exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R = 0.3426 * 298.15 \ K$

$-a_{12}/R =102.15 \ K$

$a_{12}/R = -102.15 \ K$

From equation (1) ; let replace 178.98 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = $\frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )$

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = $\frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )$

∧₂₁ = 0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

8 0

3 years ago

Which equation is balanced? “Hint...count the atoms of each element. Only compare like elements!

lisabon 2012 [21]

Answer:

2Mg + O2 = 2MgO

Explanation:

Al + 3F2 = AlF3 ..... F has 6 on the left and 3 on the right.

Na + F2 = 2NaF ..... Na has 1 on the left and 2 on the right.

2Al + 3O2 = Al2O3 .... O has 6 on the left and 3 on right.

8 0

3 years ago