All categories

3 years ago

Compared to the boiling point and the freezing point of water at 1 atmosphere, a 1.0 M CaCl2(aq) solution at 1 atmosphere has a

Chemistry

2 answers:

loris [4]3 years ago

6 0

Answer:

The is a higher boiling

point and lower freezing point

Explanation:

Vlada [557]3 years ago

4 0

Answer:

Higher boiling point and lower freezing point

Explanation:

Boiling point is the temperature at which substances goes to gaseous phase form their liquid phase. Liquids boil when vapour pressure of the liquid becomes equal to atmospheric pressure. So boiling point of a liquid is affected by the change in its vapour pressure and vapour pressure is affected by adding a non-volatile solute in it.

Effect of adding non -volatile solute on vapour pressure:

When a non-volatile solute is added to a liquid, it alters the interaction between the liquid particles and prevents the liquid particles to going into gaseous phase which results in decrease in vapour pressure. More energy is needed for the vapour pressure of the liquid to becomes equal to atmospheric pressure. Therefore, boiling point increases.

CaCl2 is a non-volatile solute and hence increases the boiling point of the water.

Effect of adding non-volatile solute on freezing point:

When a non-volatile solute is added to a liquid, it alters the interaction between the liquid molecules. The new solute and solvent interaction prevents the liquid to soldify and requires more lowering in temperature. Therefore, freezing point decreases.

You might be interested in

How much ice (in grams) would have to melt to lower the temperature of 353 mL of water from 26 ∘C to 6 ∘C? (Assume the density o

Rashid [163]

Answer:

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

Explanation:

Heat gain by ice = Heat lost by water

Thus,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=-m_{water}\times C_{water}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

Heat of fusion + $m_{ice}\times C_{ice}\times (T_f-T_i)=m_{water}\times C_{water}\times (T_i-T_f)$

Heat of fusion = 334 J/g

Heat of fusion of ice with mass x = 334x J/g

For ice:

Mass = x g

Initial temperature = 0 °C

Final temperature = 6 °C

Specific heat of ice = 1.996 J/g°C

For water:

Volume = 353 mL

$Density (\rho)=\frac{Mass(m)}{Volume(V)}$

Density of water = 1.0 g/mL

So, mass of water = 353 g

Initial temperature = 26 °C

Final temperature = 6 °C

Specific heat of water = 4.186 J/g°C

So,

$334x+x\times 1.996\times (6-0)=353\times 4.186\times (26-6)$

$334x+x\times 11.976=29553.16$

345.976x = 29553.16

x = 85.4197 kg

Thus,

The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg

7 0

4 years ago

How many moles are in 2.4g of carbon dioxide (CO2)?

miv72 [106K]

Answer:

$\boxed {\boxed {\sf 0.055 \ mol \ CO_2}}$

Explanation:

To convert form grams to moles, the molar mass must be used. This is the mass (in grams) in 1 mole of a substance.

We can use the values on the Periodic Table. First, find the molar masses of the individual elements: carbon and oxygen.

C: 12.011 g/mol
O: 15.999 g/mol

Check for subscripts. The subscript of 2 after O means there are 2 oxygen atoms, so we have to multiply oxygen's molar mass by 2 before adding.

O₂: 2* (15.999 g/mol)=31.998 g/mol
CO₂: 12.011 g/mol + 31.998 g/mol =40.009 g/mol

Use the molar mass as a ratio.

$\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}$

Multiply by the given number of grams.

$2.4 \ g \ CO_2 *\frac {44.009 \ g\ CO_2}{ 1 \ mol \ CO_2}$

Flip the fraction so the grams of carbon dioxide cancel.

$2.4 \ g \ CO_2 *\frac { 1 \ mol \ CO_2}{44.009 \ g\ CO_2}$

$2.4 *\frac { 1 \ mol \ CO_2}{44.009}$

$\frac { 2.4 \ mol \ CO_2}{44.009}= 0.0545342998 \ mol \ CO_2$

The original measurement of grams has 2 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

The ten thousandth place has a 5, so we round the 4 to a 5.

$0.055 \ mol \ CO_2$

2.4 grams of carbon dioxide is about 0.055 moles.

8 0

3 years ago

What are three elements that have similar chemical properties to oxygen

jok3333 [9.3K]

Sulfur, selenium, and tellurium

4 0

3 years ago

The natural distribution of the isotopes of a hypothetical element is 60.795% at a mass of 281.99481 u, 22.122% at a mass of 283

fenix001 [56]

Answer: The average atomic mass of the element is 283.291 amu

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1:

Mass of isotope 1 = 281.99481 amu

Percentage abundance of isotope 1 = 60.795 %

Fractional abundance of isotope 1 = 0.60795

For isotope 2:

Mass of isotope 2 = 283.99570 amu

Percentage abundance of isotope 2 = 22.122 %

Fractional abundance of isotope 2 = 0.22122

For isotope 3:

Mass of isotope 3 = 286.99423 amu

Percentage abundance of isotope 3 = [100 - (60.795 + 22.122)] = 17.083 %

Fractional abundance of isotope 1 = 0.17083

Putting values in equation 1, we get:

$\text{Average atomic mass of element}=[(281.99481\times 0.60795)+(283.99570\times 0.22122)+(286.99423\times 0.17083)]\\\\\text{Average atomic mass of element}=283.291amu$

Hence, the average atomic mass of the element is 283.291 amu

5 0

4 years ago

Which option is an element?

insens350 [35]

Answer:

Explanation:

water

3 0

3 years ago

Read 2 more answers