What is the valency of water

Chemistry

1 answer:

just olya [345]2 years ago

3 0

Answer:

In the water molecule oxygen combines with two hydrogens and so has a valency of 2. A simple diagram of a water molecule makes this plain. The hydrogen atoms are each sharing one pair of electrons - they have a valency of 1. The oxygen atom is sharing two pairs of electrons - it has a valency of 2.

You might be interested in

Lithium is a metal,chlorine is a non metal,so LiCl would be a ______bond

ivann1987 [24]

Ionic is the answer. This is because lithium has a positive charge, while chlorine has a negative charge, meaning the compound doesn’t necessarily have an overall charge.

4 0

3 years ago

Explain how you determine the freezing point of a solution that does not have a well-defined transition in the cooling curve.

Sophie [7]

This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

$y=-3.5 x + 25\\\\y=-0.52 x + 2$

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

$-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72$

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

$y=-3.5 (7.72) + 25\\\\y = 1.84$

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.

Learn more: