What type(s) of bond(s) does carbon have a tendency to form?

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction At

anygoal [31]

Answer: [N2]₀ = 10M and [H2]₀ = 11M

Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:

I is initial amount;

C is change in concentration;

E is for equilibrium concentration;

For the mixture,

N2 3H2 2NH3

I [N2]₀ [H2]₀ 0

C - x -3x +2x

E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4

With the product, we can find "x":

2x=4

x=2M

With x=2, find the concentrations:

[N2]₀ - x = 8

[N2]₀ = 10M

[H2]₀ - 3x = 5

[H2]₀ = 11M

The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.

8 0

3 years ago

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C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

3 0

3 years ago

This question has two parts btw

Oxana [17]

Answer:

the cold one is going to have a faster reaction time

Explanation:

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3 years ago

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If the two atoms participating in a bond have an electron negativity difference of 1.9 , what type of of bond would they be part

brilliants [131]

Answer:

Polar covalent bond.

Explanation:

When the bond is formed between the atoms by sharing the electrons the bond thus have covalent character. The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive. When the electronegativity difference is less than 0.4 the bond is non polar covalent.

When bonded atoms have greater electronegativity difference i.e 2 or greater than two the bond is ionic because electron is transfer from low electronegative atom to highest electronegative atom.

For example:

In water the electronegativity of oxygen is 3.44 and hydrogen is 2.2. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and hydrogen becomes partial positive.

In case of H₂, Cl₂, Br₂ the bond has very high covalent character because of zero electronegativity difference.

7 0

3 years ago

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n =

Lady bird [3.3K]

Answer:

$4.86\times10^{-7}\ \text{m}$

Explanation:

R = Rydberg constant = $1.09677583\times 10^7\ \text{m}^{-1}$

$n_1$ = Principal quantum number of an energy level = 2

$n_2$ = Principal quantum number of an energy level for the atomic electron transition = 4

Wavelength is given by the Rydberg formula

$\lambda^{-1}=R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\\\Rightarrow \lambda^{-1}=1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\\\Rightarrow \lambda=\left(1.09677583\times 10^7\left(\dfrac{1}{2^2}-\dfrac{1}{4^2}\right)\right)^{-1}\\\Rightarrow \lambda=4.86\times10^{-7}\ \text{m}$

The wavelength of the light emitted is $4.86\times10^{-7}\ \text{m}$ .

6 0

3 years ago