What causes atoms to form covalent bonds?

Do anyone know how to do question B

Over [174]

Answer:

a) IUPAC Names:

1) (<em>trans</em>)-but-2-ene

2) (<em>cis</em>)-but-2-ene

3) but-1-ene

b) Balance Equation:

C₄H₁₀O + H₃PO₄ → C₄H₈ + H₂O + H₃PO₄

As H₃PO₄ is catalyst and remains unchanged so we can also write as,

C₄H₁₀O → C₄H₈ + H₂O

c) Rule:

When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.

d) C is not Geometrical Isomer:

For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.

7 0

3 years ago

N2(g) + 2H(g) -> N2 H4(g) What are the volumes of N2 gas and H2 gas required to form 28.5 grams of N2 H4 at 30'C and 1.50 atm

FromTheMoon [43]

Answer:

Volume of N₂ = 14.76 L

Volume of H₂ = 29.52 L

Explanation:

Given data:

Mass of N₂H₄ formed = 28.5 g

Pressure = 1.50 atm

Temperature = 30°C (30+273 = 303 k)

Volume of N₂ and H₂ needed = ?

Solution:

Chemical equation:

N₂ + 2H₂ → N₂H₄

Number of moles of N₂H₄ formed = mass/ molar mass

Number of moles of N₂H₄ formed = 28.5 g/ 32 g/mol

Number of moles of N₂H₄ formed = 0.89 mol

Now we will compare the moles of N₂H₄ with N₂ and H₂ form balance chemical equation.

N₂H₄ : N₂

1 : 1

0.89 : 0.89

N₂H₄ : H₂

1 : 2

0.89 : 2×0.89 = 1.78 mol

Volume of H₂:

PV = nRT

1.50 atm × V = 1.78 mol × 0.0821 atm.L/mol.K × 303 K

V = 44.28atm.L /1.50 atm

V = 29.52 L

Volume of N₂:

PV = nRT

1.50 atm × V = 0.89 mol × 0.0821 atm.L/mol.K × 303 K

V = 22.14 atm.L /1.50 atm

V = 14.76 L

6 0

3 years ago

Cierta cantidad de gas ocupa 20 litros sometido a una presión de 1,5 atmósferas. Si lo comprimimos hasta 6 atmósferas, ¿qué volu

Mekhanik [1.2K]

Respuesta:

5 L

Explicación:

Paso 1: Información provista

Presión inicial (P₁): 1,5 atm

Volumen inicial (V₁): 20 L

Presión final (P₂): 6 atm

Paso 2: Calcular el volumen final del gas

Si asumimos temperatura constante y comportamiento ideal, podemos calcular el volumen final del gas (V₂) usando la Ley de Boyle.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 1,5 atm × 20 L / 6 atm = 5 L

4 0

3 years ago

Ka/KbMIXED PRACTICE108.Calculate the [H3O+(aq)], the pH, and the % reaction for a 0.50 mol/L HCN solution. ([H3O+(aq)] = 1.8 x 1

NeTakaya

Answer:

a) [H₃O⁺] = 1.8x10⁻⁵ M

b) pH = 4.75

c) % rxn = 3.5x10⁻³ %

Explanation:

a) The dissociation reaction of HCN is:

HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)

0.5 M - x x x

The dissociation constant from the above reactions is given by:

$Ka = \frac{[H_{3}O^{+}][CN^{-}]}{[HCN]} = 6.17 \cdot 10^{-10}$

$6.17 \cdot 10^{-10} = \frac{x*x}{(0.5 - x)}$

$6.17 \cdot 10^{-10}*(0.5 - x) - x^{2} = 0$

By solving the above quadratic equation we have:

x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]

Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.

b) The pH is equal to:

$pH = -log[H_{3}O^{+}] = -log(1.75 \cdot 10^{-5} M) = 4.75$

Then, the pH of the HCN solution is 4.75.

c) The % reaction is the % ionization:

$\% = \frac{x}{[HCN]} \times 100$

$\% = \frac{1.75 \cdot 10^{-5} M}{0.5 M} \times 100$

$\% = 3.5 \cdot 10^{-3} \%$

Therefore, the % reaction or % ionization is 3.5x10⁻³ %.

I hope it helps you!

6 0

3 years ago

What are the ways humans use, collect, and distribute renewable and nonrenewable natural resources?

guajiro [1.7K]

Answer:

Air and water pollution, damage to public health, wildlife and habitat loss, water use, land use, and global warming emissions.

<h2>Explanation: <u><em>HOPE THIS HELPS!!!!</em></u></h2>

8 0

2 years ago