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Leona [35]
3 years ago
13

Water molecules have a polarity, which allows them to be electrically attracted to other water molecules and other polar molecul

es by weak chemical bonds known as _____. View available hint(s)

Chemistry
1 answer:
-BARSIC- [3]3 years ago
4 0
<h3>Answer:</h3>

            Water molecules have a polarity, which allows them to be electrically attracted to other water molecules and other polar molecules by weak chemical bonds known as <u>Hydrogen Bond</u>.

<h3>Explanation:</h3>

                     Hydrogen Bond Interactions are those interactions which are formed between a <em>partial +ive</em> hydrogen atom bonded directly to most electronegative atoms (i.e. Fluorine, Oxygen and Nitrogen) of one molecule interacts with the <em>partial -ive</em> most electronegative atom of another molecule. These interactions are considered the strongest intermolecular interactions and are responsible for many stronger physical properties of water like, boiling point, viscosity, solubility e.t.c.

                       Attached figure shows the hydrogen bond interactions (green dashed lines) between water molecules. The partial charges are shown in red color.

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A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o
julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

Number \, of \, moles, n  = \frac{Mass}{Molar \ mass} =  \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0
3 years ago
2CuS + 3O2 = 2CuO + 2SO2
Nata [24]
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3 years ago
The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol
Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration PCl_3 and Cl_2.

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}

\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M

\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}

\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M

The given equilibrium reaction is,

                            PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

Initially                 0.70        0.70              0

At equilibrium    (0.70-x)   (0.70-x)           x

The expression of K_c will be,

K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}

K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}

Now put all the given values in the above expression, we get:

49=\frac{(x)}{(0.70-x)\times (0.70-x)}

By solving the term x, we get

x=0.59\text{ and }0.83

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of PCl_3 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of Cl_2 at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of PCl_5 at equilibrium = x = 0.59 M

Therefore, the concentration of PCl_3 at equilibrium is 0.11 M

3 0
3 years ago
The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
3 years ago
An atom has the following electron configuration 1s2 2s2 2p6 3s2 3p4 . How many valence electrons does this Atom have
yawa3891 [41]

Answer:

6

Explanation:

This atom is sulfur (if the electrons are equal to the protons/not an ion). You can tell the number of valence electrons by looking at the individual shell. The first shell (1s) can only hold 2 electrons. The second shell (2s and 2p) can hold 8 electrons. The third shell (3s and 3p), which is the valence shell, only has 6 out of its possible 8 electrons, so this atom has 6 valence electrons.

6 0
3 years ago
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