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3 years ago

In which shell are the valence electrons of the elements in Period 2 found?

Chemistry

2 answers:

defon3 years ago

8 0

B 2
is which shell are the valence electrons of the elements in period 2 found

Diano4ka-milaya [45]3 years ago

4 0

Answer:

(2) 2

Explanation:

The period of an element indicates in which shell are the valence electrons. So, all the elements of the Period 2 have their valence electrons in the 2nd shell.

A period in the periodic table is a row of chemical elements.

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Complete combustion of a 0.600-g sample of a compound in a bomb calorimeter releases 24.0 kJ of heat. The bomb calorimeter has a

coldgirl [10]

The final temperature, t₂ = 30.9 °C

<h3>Further explanation</h3>

Given

24.0 kJ of heat = 24,000 J

Mass of calorimeter = 1.3 kg = 1300 g

Cs = 3.41 J/g°C

t₁= 25.5 °C

Required

The final temperature, t₂

Solution

Q = m.Cs.Δt

Q out (combustion of compound) = Q in (calorimeter)

24,000 = 1300 x 3.41 x (t₂-25.5)

t₂ = 30.9 °C

3 0

3 years ago

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Where are the rare earth metals located on the periodic table

maksim [4K]

They are found in group 3 of the periodic table.

7 0

3 years ago

The electron cloud model describes the __of electrons in an atom.

Alex Ar [27]

Answer:

it's location

Explanation:

4 0

3 years ago

Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

6 0

3 years ago

How many electrons are in the highest occupied energy level of a nuetral silicon atom?

Ludmilka [50]

A). 4
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3 years ago