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2 years ago

Help pleased alot of points

Chemistry

1 answer:

JulsSmile [24]2 years ago

4 0

Answer:

Nitrogen non metal

Gain 3 electrons

Negative ion N3-

Barium metal

Lose electrons

Positive charge

Selinium non metal

Gain electron

Negative charge

Cesium metal

Lose electrons

Positive charge

Explanation:

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The first step in the free radical mechanism for the preparation of polyethylene is:

kherson [118]

Answer:

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

Explanation:

The preparation of ethylene occur by free radical mechanism , which occurs in three steps :

1. Chain Initiation

2. Chain propagation

3. Chain termination.

The first step i.e chain initiation require a radical. The compound are generally peroxide(O-O linkage) that produce radical is :

Benzoyl Peroxide

This compound when heated (with U.V radiation) produce two species of radicals .

(see the attached image)

These RADICALS are highly unstable and very reactive in nature.

They readily react with ethene and initiate the polymerisation.

Thus,

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

<u><em>The sequence of Polymerisations are:</em></u>

b. heating an organic peroxide to break the O-O bond

d. propagation of the free radicals

3 0

3 years ago

2.1 g of a hydrocarbon fuel is burned in a calorimeter that contains 280 grams of water initially at 25.00◦C. After the combusti

olga2289 [7]

Answer:

$\large \boxed{\text{933 J}}$

Explanation:

There are three heat transfers involved.

heat from combustion of propane + heat gained by water + heat gained by calorimeter = 0

q₁ + q₂ + q₃ = 0

m₁ΔH + m₂C₂ΔT + C_calΔT = 0

Data:

m₁ = 2.1 g

m₂ = 280 g

Ti = 25.00 °C

T_f = 26.55 °C

Ccal = 92.3 J·°C⁻¹

Calculations:

Let's calculate the heats separately.

1. q₁

q₁ = 2.1 g × ΔH = 2.1ΔH g

2. q₂

ΔT = T_f - Ti = 26.55 °C - 25.00 °C = 1.55 °C

q₂ = 280 g × 4.184 J·°C⁻¹ × 1.55 °C = 1816 J

3. q₃

q₃ = 92.3 J·°C⁻¹ × 1.55 °C = 143.1 J

4. ΔH

$\begin{array}{rcl}\text{2.1$\Delta$H g +1816 J +143.1 J} & = & 0\\\text{2.1$\Delta$H g +1959 J} & = & 0\\\text{2.1$\Delta$H g}& = & \text{-1959 J}\\\Delta H & = & \dfrac{\text{-1959 J}}{\text{2.1 g}}\\\\& = & \textbf{-933 J/g}\\\end{array}\\\text{The combustion releases $\large \boxed{\textbf{933 J}}$ per gram of fuel burned.}$