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3 years ago

What is meant by the term polar molecule

Chemistry

2 answers:

alukav5142 [94]3 years ago

4 0

An asymetric molecule with non uniform positive an negative charges

s344n2d4d5 [400]3 years ago

3 0

Separation of electric charge from one molecule to another<span />

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Which is an acceptable Lewis structure for a diatomic nitrogen molecule?

madam [21]

Https://quizlet.com/18404690/chemistry-final-flash-cards/

here is a Quizlet I have for that

7 0

4 years ago

Find the partial pressure of helium gas in mmHg

mezya [45]

Answer:

214 mmhg

Explanation:

5 0

2 years ago

A 15.0 g sample of nickel metal is heated to 100.0 degrees C and dropped into 55.0 g of water, initially at 23.0 degrees C. Assu

OLEGan [10]

Answer: The final temperature of nickel and water is $25.2^{o}C$ .

Explanation:

The given data is as follows.

Mass of water, m = 55.0 g,

Initial temp, $(t_{i}) = 23^{o}C$ ,

Final temp, $(t_{f})$ = ?,

Specific heat of water = 4.184 $J/g^{o}C$ ,

Now, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$

Also,

mass of Ni, m = 15.0 g,

Initial temperature, $t_{i} = 100^{o}C$ ,

Final temperature, $t_{f}$ = ?

Specific heat of nickel = 0.444 $J/g^{o}C$

Hence, we will calculate the heat energy as follows.

q = $mS \Delta t$

= $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$

Therefore, heat energy lost by the alloy is equal to the heat energy gained by the water.

$q_{water}(gain) = -q_{alloy}(lost)$

$55.0 g \times 4.184 J/g^{o}C \times (t_{f} - 23^{o}C)$ = -( $15.0 g \times 0.444 J/g^{o}C \times (t_{f} - 100^{o}C)$ )

$t_{f} = \frac{25.9^{o}C}{1.029}$

= $25.2^{o}C$

Thus, we can conclude that the final temperature of nickel and water is $25.2^{o}C$ .

6 0

3 years ago

ch question carries 2 mark. Time Remaining : 00 : 46 : 33 Some oxides are given below. (i)Na2O (ii)NO2 (iii) CO2 (iv) MgO a) Whi

cluponka [151]

<h3>Further explanation</h3>

The basic oxide is an oxide-forming a base solution.

These oxides are mainly from group 1 alkaline and group 2-alkaline earth

If this oxide is dissolved in water it will form an alkaline solution

LO + H₂O --> L(OH)₂ ---> alkaline earth

L₂O + H₂O --> LOH --> alkaline

So the basic oxides : Na₂O and MgO

Na₂O + H₂O --> NaOH (sodium hydroxide, strong base)

MgO + H₂O --> Mg(OH)₂ (magnesium hydroxide, strong base)

The aqueous solution of CO₂ , obtained by dissolving CO₂ in water

CO₂ + H₂O --> H₂CO₃ (carbonic acid)

In general, basic oxide is obtained from metal oxide, while acid oxide is obtained from non-metal oxide

6 0

4 years ago

Red light has a wavelength of about 6.5 x 10-7m. Find its frequency. Show all work.

ozzi

For this problem we use the wave equation. It is expressed as the speed (c) is equal to the product of frequency (f) and wavelength (v).

c = v x f

We know the wavelength of the an red light which is 6.5 x 10^-7 m. Now, we solve for the wavelength of the unknown wave to see the relation between the two waves.

2.998 X 10^8 = 5.3 X 10^15 X v
v = 2.998 X 10^8 / (5.3 X 10^15) = 5.657 X 10^-8 m

Therefore, the wavelength of the unknown wave is less than the wavelength of the red light.

3 0

3 years ago