Solutions that can conduct electricity are called .

Which way would CCI2F2 orient in an electric field?

ArbitrLikvidat [17]

Answer:

See explanation

Explanation:

In looking at molecules to determine whether they are polar or not we have to look at two things basically;

i) presence of polar bonds

ii) geometry of the molecule

Now, we know that CCI2F2 is a tetrahedral molecule, but the molecule is not symmetrical. It has four polar bonds that are not all the same hence the molecule is polar.

In an electric field, polar molecules orient themselves in such a way that the positive ends of the molecule are being attracted to the negative plate while the negative ends of the molecules are attracted to the positive plate.

So the positive ends of CCI2F2 are oriented towards the negative plate of the field while the negative ends of CCI2F2 are oriented towards the positive ends of the field.

3 0

3 years ago

the white powder is added to the clear liquid. what event is occurring and is it a physical or chemical change?

alexdok [17]

I would say chemical but I’m not 100% sure might wanna get a second opinion

7 0

4 years ago

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Oxyacetylene torches produce such high temperature that they are often used to weld and cut metal. When 1.53 g of acetylene (C2H

hodyreva [135]

Answer: $4403kJ/mole$

Explanation:

First we have to calculate the heat absorbed by bomb calorimeter

Formula used :

$q_b=c_b\times (T_{final}-T_{initial})$

$q_b$ = heat absorbed by calorimeter = ?

$c_b$ = specific heat of = 10.69 kJ/K

$T_{final}$ = final temperature = $44.688^oC=(273+44.688)K=317.688K$

$T_{initial}$ = initial temperature = $20.486^oC=(273+20.486)K=293.486K$

$q_b=10.69kJ/K\times (317.688-293.486)=258.7kJ$

As heat absorbed by calorimeter is equal to the heat released by acetylene during combustion.

Thus 1.53 gram of acetylene releases heat of combustion = 258.7kJ

So, 26.04 g/mole of acetylene releases heat of combustion $\frac{258.7}{1.53}\times 26.04=4403kJ/mole$

Therefore, the heat of combustion of acetylene is, $4403kJ/mole$

7 0

4 years ago

How many moles of water (H2O) are present in a beaker containing 45.9 g H2O? Give your answer to the correct number of significa

charle [14.2K]

Explanation:

Number of moles is defined as the mass of given sample divided by the molar mass of sample.

Mathematically, No. of moles = $\frac{mass given}{Molar mass}$

Since, it is given that mass of water is 45.9 g and it is known that molar mass of water is 18.02 g/mol.

Therefore, calculate the number of moles as follows.

No. of moles = $\frac{mass given}{Molar mass}$

= $\frac{45.9 g}{18.02 g/mol}$

= 2.547 mol

Thus, we can conclude that there are 2.547 mol present in 45.9 g of water.

7 0

4 years ago

Read 2 more answers

Calculate the work (kJ) done during a reaction in which the internal volume expands from 20 L to 43 L against an outside pressur

Alenkinab [10]

Answer:

$\large \boxed{\text{-10.0 kJ}}$

Explanation:

1. Calculate the work

w = - pΔV = -4.3 atm × (43 L - 20 L) = -4.3 × 23 L·atm = -98.9 L·atm

2. Convert litre-atmospheres to joules

$w = \text{-98.9 L\cdot$atm } \times \dfrac{\text{101.3 J}}{\text{1 L$\cdot$atm }} = \text{-10000 J} = \textbf{-10.0 kJ}\\\\\text{The work done is $\large \boxed{\textbf{-10.0 kJ}}$}$

The negative sign indicates that the work was done against the surroundings.

4 0

3 years ago