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mario62 [17]
4 years ago
8

Which type of orbitals overlap to form the sigma bond between c and cl in ch3cl?

Chemistry
1 answer:
____ [38]4 years ago
3 0
<span>The hybridization between c and cl in ch3cl is sp3-sp3 because it will maximize the electron distribution in cl.in ch3cl carbon is considered to be sp3 hybridized.the ch3cl molecule is assumed to be tetrahydral in shape with bond angles in the neighborhoodof 109.5 degrees.The cl has sp3 hybridization due to the four pairs of electrons.</span>
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A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is
aliina [53]

Answer:

\large \boxed{\text{67 000 g}}

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂  = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

 m = the mass of the sample

  C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}

7 0
3 years ago
Which of the following will increase the number of collisions without changing the energy of reactant molecules?
natulia [17]

Answer:increasing the concentration of reactants

Explanation:

Collision is the phenomenon in which the reactant molecules come to nearest closness,as a result the reactants are converted into products.

Now the number of effective collision is directly proportional to the number of reactants added..

6 0
3 years ago
At stp what is the volume of 5.35 moles of methane ch4​
Nookie1986 [14]

Answer:

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Explanation:

8 0
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What’s the question?
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An equilibrium mixture of O2, SO2 and SO3 contains equal concentrations of SO2 and SO3.
AleksAgata [21]

Answer:

[O₂(g)] = 0.0037M

Explanation:

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Conc:   [SO₂(g)]   [O₂(g)]     [SO₃(g)]  and  [SO₂(g)] = [SO₃(g)]

Kc =  [SO₃(g)]²/[O₂(g)][SO₂(g)]² => Kc = 1/[O₂(g)] = 270 if [SO₂(g)] = [SO₃(g)]

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7 0
3 years ago
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