A chemical reaction in which compounds break up into simpler constituents is a ______

For a 0.50 molal solution of sucrose in water, calculate the freezing point and the boiling point of the solution.

jok3333 [9.3K]

Answer:

-0.93 °C; 100.26 °C

Step-by-step explanation:

(a) Freezing point depression

The formula for the freezing point depression ΔT_f is

ΔT_f = iKf·b

i is the van’t Hoff factor: the number of moles of particles you get from a solute.

For sucrose,

Sucrose (s) ⟶ sucrose (aq)

1 mole sucrose ⟶ 1 mol particles i = 1

ΔT_f = 1 × 1.86 × 0.50

ΔT_f = 0.93 °C

T_f = T_f° - ΔT_f

T_f = 0.00 – 0.93

T_f = -0.93 °C

(b) Boiling point elevation

The formula for the boiling point elevation ΔTb is

ΔTb = iKb·b

ΔTb = 1 × 0.512 × 0.50

ΔTb = 0.256 °C

Tb = Tb° + ΔTb

Tb = 100.00 + 0.256

Tb = 100.26 °C

6 0

2 years ago

what is the approximate ph at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous formic acid require

vivado [14]

Answer:

pH at equivalence point is 8.52

Explanation:

$HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O$

1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of $HCOO^{-}$

So, moles of NaOH used to reach equivalence point equal to number of moles $HCOO^{-}$ produced at equivalence point.

As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.

So, moles of $HCOO^{-}$ produced = $\frac{29.80\times 0.3567}{1000}moles=0.01063moles$

Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL

So, at equivalence point concentration of $HCOO^{-}$ = $\frac{0.01063\times 1000}{54.80}M=0.1940M$

At equivalence point, pH depends upon hydrolysis of $HCOO^{-}$ . So, we have to construct an ICE table.

$HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}$

I: 0.1940 0 0

C: -x +x +x

E: 0.1940-x x x

So, $\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}$

species inside third bracket represent equilibrium concentrations

So, $\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}$

or, $x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0$

So, $x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}$

So, $x=3.285\times 10^{-6}M$

So, $pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52$

5 0

2 years ago

Describe how gasoline, as a fuel, can undergo a chemical change.

kvasek [131]

Answer with Explanation:

A "chemical change" occurs when the structure of the molecules change. If there's no change in structure, it is called a "physical change." This also means that a chemical reaction has to occur for something to be considered a chemical change.

Gasoline when used as a fuel is being transformed into other forms of energy. It undergoes a process called "combustion."

When used as a fuel for cars, the gasoline is being burned to produce heat energy. This energy is then converted into mechanical energy that allows the car to move. Clearly, the gasoline has undergone a chemical change.

5 0

2 years ago

What are some chemacial changes in your school?

Oksanka [162]

Putting glue on something, because once it is set in you cannot change it back.

6 0

3 years ago

What is the diffrence between a mixture and a compound

love history [14]

A compound is a substance made up of atoms of two or more different elements joined by chemical bonds. A mixture is a combination of two or more substances that are not chemically combined.

6 0

3 years ago

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A chemical reaction in which compounds break up into simpler constituents is a _______ reaction.