Chemistry, 07.12.2021 19:40 brighamc23

Given the following reaction Ca(OH)2+H2SO4→CaSO4+2H2O
If you start with 14.82 g of Ca(OH)2 and 16.35 g of H2SO4,
a) determine the limiting reagent
b) determine the number of moles of H2O produced
c) determine the number of grams of CaSO4 produced
d) determine the number of grams of excess reagent left

Answers: 1

Show answers

Answers

Answer from: Quest

Standing in the sunlight and feeling the heat on your face (uv radiation is entering the skin, converting to infrared radiation, and is absorbed by your tissues)orstanding near a fire and feeling the heat on your skin (infrared radiation transfer)

Answer from: Quest

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

Answer from: Quest

the term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. hydride compounds in general form with almost any element, except a few noble gases. the trends and properties vary according to the type of intermolecular force that bonds the elements together, the temperature, its molecular masses, and other components. hydrides are classified into three major groups, depending on what elements the hydrogen bonds to. the three major groups are covalent, ionic, and metallic hydrides. formally, hydride is known as the negative ion of a hydrogen, h-, also called a hydride ion. because of this negative charge, hydrides have reducing, or basic properties. its special characteristics will be further discussed.

explanation:

Answer from: Quest

Hello,#agreed with joseph002.i got my source also from source: hope this