Chemistry, 12.08.2021 02:00, greenclonetroop6q2tg

Encuentra predicado, verbo, objeto directo y objeto indirecto en las siguientes oraciones: 1) Hola, vengo a devolverte los guantes que me prestaste.
2) Reservemos dos entradas para Juan y Alberto.
3) La lluvia cambió los planes para el resto de la jornada.
4) Nunca le agradeciste sus buenos consejos.
5) Describen la fiesta a los amigos que no pudieron ir.
6) En su enojo, le arrojó el agua del vaso a la cara
7) Llevó postres para los invitados.
8) Compre la casa para mis hijos.

Answers: 2

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answered: Guest

i think it's letter a.

answered: Guest

It could possibly be beneficial if the experiment is successful.

answered: Guest

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

$m = \frac{moles (solute)}{l ( solution)}$

$1 m = \frac{moles (cacl_{2})}{1 l }= 1 mol cacl_{2}$

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ = $1 mol \times \frac{111g}{mol} = 111 g cacl_{2}$

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution = $\frac{1g}{ml} \times 1000 ml = 1000 g$

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

$molality = \frac{mol (solute)}{kg ( solvent)}$

$molality = \frac{1mol}{0.889kg}$

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

$\bigtriangleup t_{b} = i \times k_{b} \times m$

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as $cacl_{2} \rightarrow ca^{2+} + 2 cl^{-}$

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

$\bigtriangleup t_{b} = 3 \times 0.512 \times 1.12$

$\bigtriangleup t_{b} = 1.7$

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

$\bigtriangleup t_{f} = i \times k_{f} \times m$

kf is constant for water and its value is 1.86

$\bigtriangleup t_{f} = 3 \times 1.86 \times 1.12$

$\bigtriangleup t_{f} = 6.2$

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point

answered: Guest

First up write the aim
then write the materials reqd.
then the procedure which you follow
then observation
then conclusion
then you can write lab safety rules or precautions(you can avoid if u want).
lemme' give an example
aim: to observe the double displacement reaction
materials reqd: bacl2 and na2so4, test tubes and a dropper
procedure:
1. take bacl2 in a test tube and na2so4 in the other.
2. then pour bacl2 into na2so4 solution.
3. observe
observation:
you will see that a white precipitate is formed called baso4. along with that common salt or nacl is formed.
the equation is
bacl2+na2so4=baso4+2nacl
conclusion:
a double displacement reaction is observed.
hope it