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bagirrra123 [75]

3 years ago

13

H2O consists of polar molecules. CO2 consists of nonpolar molecules. How do chemists explain this difference?

2 answers:

Vadim26 [7]3 years ago

8 0

C. Oxygen has a different attraction for electrons in H2O than in CO2

the covalent bond between Hydrogen-Oxygen and Carbon-Oxygen is different because the difference of electronegativity (the tendency of an atom to attract a shared pair of electrons towards itself) between Hydrogen and Oxygen is a lot higher than the difference of electronegativity between Carbon and Oxygen. This difference in electronegativity results in separation of partial electric charge creating polar bonds and giving the molecule a polarity.

lesantik [10]3 years ago

8 0

Answer:

These compounds have different shapes.

Explanation:

Two things determine whether a molecule is polar or no polar. The size and shape of the molecule.

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A 1.00-kg object is attached by a thread of negligible mass, which passes over a pulley of negligible mass, to a 2.00-kg object.

Anna [14]

Answer:

a = 3.27 m/s²

v = 2.56 m/s

Explanation:

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mass B = 2 kg

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3 years ago

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A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago