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3 years ago

Which of these solutes raises the boiling point of water the most?

1 answer:

4 0

Raising of the boiling point is a colligative property. That means that it depends on the number of particles dissolved. The greater the number of particles the greater the increase in the boiling point. So, you can compare the effect of these solutes in the increase of the boiling point by writing the chemical equations and comparing the number of particles dissolved: 1)ionic lithium chloride, LiCl(s) --> Li(+) + Cl (-) => 2 ions; 2) ionic sodium chloride, NaCl(s) --> Na(+) + Cl(-) => 2 ions; 3) molecular sucrose, C12H22O11 (s) ---> C12H22O11(aq) => 1 molecule; 4) ionic phosphate, Na3PO4 --> 3Na(+) + PO4 (3-) => 4 ions; 5) ionic magnesium bromide, MgBr2 --> Mg(2+) + 2 Br(-) => 3 ions. <span>So, ionic phosphate produces the greatest number of particles and it will cause the greatest increase of the boiling point.</span><span />

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Is CO2 a covalent or ionic bond? will give brainliest for the correct answer

NeX [460]

Answer:

covalent bond

Explanation:

a covalent bond happens between two nonmetals

a ionic bond happens between a metal and a non metal

CO2 is a bond between carbon and oxygen

carbon and oxygen are both non metals therefore CO2 is a covalent bond

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2 years ago

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In the following reaction, C6H6)? 2C6H6 + 15O2 12CO2 + 6H2O,

shepuryov [24]

The answer is 27.92g . working is shown in the picture above.

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3 years ago

What do calcium chloride, sodium carbonate, calcium carbonate, and potassium carbonate all have in common? (other than they reac

VARVARA [1.3K]

Answer:

the all dissolve in water

Explanation:

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3 years ago

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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )

yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution : Given,

Concentration (c) = 0.150 M

Acid dissociation constant = $k_a=4.5\times 10^{-4}$

The equilibrium reaction for dissociation of $HNO_2$ (weak acid) is,

$HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+$

initially conc. c 0 0

At eqm. $c(1-\alpha)$ $c\alpha$ $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha )$ .

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha )$ .

$4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}$

$4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2$

$0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0$

By solving the terms, we get

$\alpha=0.0533$

No we have to calculate the concentration of hydronium ion or hydrogen ion.

$[H^+]=c\alpha=0.150\times 0.0533=0.007995 M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (0.007995 M)$

$pH=2.097\approx 2.1$

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

4 0

4 years ago

A science teacher has a supply of 50% sugar solution and a supply of 80% sugar solution. How much of each solution should the te

Monica [59]

You can establish a system of two equation with two variables.

Varibles are:
V1 = volume of the 50% sugar solution
V2 = volumen of the 80% sugar solution

Equations:
Balance of sugar:

Sugar from 50% solution: 0.5*V1
Sugar from 80% solution: 0.8*V2
Sugar in the final solution (mix): 0.6 * 105 = 63

1) 0.5V1 + 0.8V2 = 63

Final volume = volume of 50% solution + volume of 80% solution

2) V1 + V2 = 105

From (2) V1 = 105 - V2

Substitue in (1)

0.5 (105 - V2) + 0.8 V2 = 63

52.5 - 0.5V2 + 0.8V2 = 63

0.3 V2 = 63 - 52.5

0.3 V2 = 10.5

V2 = 10.5/0.3
V2 = 35mL

V1 = 105 - 35 = 70 mL

Answer: 70 mL of the 50% solution and 35 mL of the 80% solution.

5 0

4 years ago