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3 years ago

Why are metallic bonds in an alkali metal relatively weak?

Chemistry

2 answers:

alexandr1967 [171]3 years ago

6 0

alkali metals contribute only a single valence Electron.

Nezavi [6.7K]3 years ago

3 0

<span>Alkali metals contribute only one valence electron</span>

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Elemental analysis of the unknown gas from part a revealed that it is 30.45% n and 69.55% o. what is the molecular formula for t

Gnesinka [82]

Assuming we have 100 g of sample

30.45/MW of N 14g = 2.175

69.55/MW of O 16g = 4.34

4.34/2.185 = 2

for every 1 mole of N we have 2 moles of O

so the empirical formula would be NO2

without having the molecular weight of the entire molecule the molecular formula can not be determined with the information in your question

6 0

3 years ago

Why is the value listed for atomic mass for each element on the periodic table not a whole number?

sweet-ann [11.9K]

A. it is an average

8 0

4 years ago

What volume of 12.0 M sulfuric acid is needed to prepare 300.0 mL of 2.50 M solution?

Zepler [3.9K]

Answer:

n=c*V

=2.50*300x10^-3

=.75 moles of hcl

v=n/c

=.75/12

=.0625 L

convert to mL

=62.5mL

4 0

4 years ago

A species with a charge of -1 contains 57 electrons and 80 neutrons. What is the identity of the species. Note: The species show

Inga [223]

In a neutral atom, the number of protons is equal to the number of electrons.

Given that the charge on the ion of the element is -1. This means that there is one electron more than the number of protons in the element.

The atomic number = Number of protons = 57 -1 = 56

Element with atomic number 56 is Barium.

Mass number of the element = Number of protons + Number of neutrons

= 56 + 80 = 136

So the identity of the species: $^{136}Ba^{-}$

3 0

3 years ago

Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)BOND

Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

$CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,$ ΔH° = ?

We are given with:

$\Delta H_{O-O}=142 kJ/mol$

$\Delta H_{O=O}=498 kJ/mol$

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{C-H}=411 kJ/mol$

$\Delta H_{C-O}=358 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH° =

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})$

$\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)$

$\Delta H^o=-794kJ$

$\Delta H^o>0$ endothermic reaction

$\Delta H^o$ exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

4 0

4 years ago