CHEMISTRY COLLEGE

Are these statements true or false? Correct the statements that are false.a ..An atom is larger than the cell of a living thing.
b.. Different kinds of elements have different kinds of atoms.
c..The protons, neutrons and electrons are different for different kinds of atoms.
d... An atom always has the same number of protons and electrons.
e... An atom always has the same number of protons and neutrons.
f.... Atoms have no mass because they are very small.
Q2 .Compare the Rutherford's model and the Bohr's model of the atom. State one similarity and one difference between them.
Q3..what observation from rutherford's gold foil experiment made him conclude that an atom has a tiny but dense nucleus that is positively charged?
Q4.. Explain why the nucleus of an atom is positively charged, while the atom is electrically neutral.
Q5...The symbols and atomic numbers of three elements are as follows:
Ne Atomic number 10
A/ Atomic number 13
K Atomic number 19
a.. identify each element from its symbol.
b.. How many protons and electrons does an atom of each element have?
c.. Draw the electronic structure of the atom of each element.

Answers

Answer 1

Answer:

Answer:

a. false

b. true

c. false

d. false

e. false

f. false

Q2: Both have electrons orbiting around the nucleus; Bohr's model is more detailed and expands on Rutherford's

Q5:

Ne = neon

Al = aluminum

K = potassium

Ne = 10 electrons/protons

Al = 13 electrons/protons

K = 19 electrons/protons

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A natural atom possesses the atomic number of 13 and a atomic mass of 27. Three electrons are lost. From what region of the atom are they lost?

Answers

Answer:

The electrons are lost from the valence shell (outermost electron shell) of the atom.

Explanation:

This is able to be inferred not only because valence electrons being lost first is a trend but also because the atom in question has actually 3 valence electrons (13-2-8 = 3).

.......state Hess law

Answers

Hess's Law of Constant Heat Summation (or just Hess's Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.

Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate. CH3CH2 C C CH3 H2O/H2SO4/HgSO4

Answers

Answer:

Draw the predominant product(s) of the following reactions including stereochemistry when it is appropriate.

CH3CH2 C C CH3 H2O/H2SO4/HgSO4

Explanation:

The given compound is: pent-2-yne.

When it reacts with water, in presence of sulphuric acid and mercuric sulphate then a ketone is formed as shown below:

This reaction is an example of nucleophilic attack of water on carbon carbon triple bond.

The general mechanism of the reaction is hsown below:

Pent-2-yne reacts with water and form 3-pentanone.

The reaction is shown below:

Final answer:

The reaction is the hydration of an alkene in an acidic environment, resulting in the formation of 2-butanol. This result is in accordance with Markovnikov's rule, which determines the position of the hydroxyl group in the resultant product.

Explanation:

The question refers to the acidity-catalyzed hydration of an alkene. In this case, you have an alkene CH3CH2 - CC - CH3 reacting in an acidic environment with water (H2O). The reactants have been exposed to H2O/H2SO4/HgSO4. In this reaction scenario, the acidic medium (H2SO4) and the water enact the role of a nucleophile and attack the alkene, thereby hydrating it.

The product of this reaction will be 2-butanol. Its formation is guided by Markovnikov's rule, which states that in the addition of a protic acid HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents. This rule is why the hydroxyl group (-OH) attaches itself to the 2nd carbon atom in the major (predominant) product.

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Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Explanation:

(a)

Initial number of moles of $NH_(3)$ and $O_(2)$ are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

$4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O$

The stoichiometric numbers are as follows.

$v_{NH_(3)}=-4$

$v_{O_(2)}=-5$

$v_(NO)=4$

$v_{H_(2)O}=6$

The total number of moles initially present -7

$\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

Initial number of moles of $H_(2)S$ and $O_(2)$ are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

$2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)$

The stoichiometric numbers are as follows.

$v_{H_(2)S}=-2$

$v_{O_(2)}=-3$

$v_{H_(2)O}=2$

$v_{SO_(2)=2$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

Initial number of moles of $NO_(2)$ , $NH_(3)$ and $N_(2)$ are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

$6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O$

The stoichiometric numbers are as follows.

$v_{NO_(2)}=-6$

$v_{NH_(3)}=-8$

$v_{N_(2)}=7$

$v_{H_(2)O}=12$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

Mole fraction of NH3: (2 - x)/(Total moles)
Mole fraction of O2: (5 - 5x/4)/(Total moles)
Mole fraction of NO: (4x)/(Total moles)
Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.

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2 N H 3 ( g ) ⟷ N 2 ( g ) + 3 H 2 ( g ) K p = 0.83 Consider your answers above, if the initial pressures for all three species is 1 atm what is the equilibrium pressure of H2? (Hint: Your quadratic will have two solutions, which one is impossible?)

Answers

The equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.

Equilibrium pressure of H2

The equilibrium pressure of H2 is calculated by creating ICE table as follows;

2 N H3 ( g ) ⟷ N2( g ) + 3H2

I: 1 1 1

C: -2x x 3x

E: 1 - 2x 1 + x 1 + 3x

$KP = ((N_2)(H_2)^3)/((NH_3)^2) \n\n0.83 = ((1 + x)(1 + 3x)^3)/((1 - 2x)^2)$

0.83(1 - 2x)² = (1 + x)(1 + 3x)³

0.83(1 - 4x + 4x²) = (1 + x)((1 + 3x)³)

0.83 - 3.32x + 3.32x² = (1 + x)((1 + 3x)³)

0.83 - 3.32x + 3.32x² = 1 + 10x + 36x² + 54x³ + 27x⁴

27x⁴ + 54x³ + 32.68x² + 13.32x + 0.17 = 0

x = -1.379 or - 0.013

Partial pressure of H2 = 1 + 3x

H2 = 1 + 3(-1.379)

H2 = -3.13 atm

H2 = 1 + 3(-0.013)

H2 = 0.96 atm

Thus, the equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.

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Atmospheric pressure arises due to the force exerted by the air above the Earth. At higher altitudes, the mass of the air above the Earth is _____ than at sea level, and atmospheric pressure therefore _____ with altitude.

Answers

Answer:

less, decreases

Explanation:

When the pressure of an atmosphere occurs because of the force exerted so at the time of the higher altitudes, the air mass i.e. above the earth should be less as the air is attracted towards surface of an earth because of the gravity and air contains the mass that shows near the surface area so automatically the air density reduced due to which the mass also decreased

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