All categories

1 month ago

Which has a lower freezing point ice or ice cream?

Chemistry

2 answers:

VMariaS [2.9K]1 month ago

8 0

-0° is the freezing temperature of ice or ice cream.

lions [2.9K]1 month ago

4 0

Ice cream has a lower freezing point in comparison to ice. This phenomenon can be explained through the freezing point depression concept, a property that relies on the number of solute and solvent particles present. Examples of colligative properties include boiling point elevation and osmotic pressure. When a nonvolatile solute, like sugar, is added to a solvent, the freezing point is lowered. In the case of ice cream, sugar acts as the nonvolatile solute, resulting in a drop in freezing temperature.

You might be interested in

If it takes 15 s for a certain sample of neon to effuse through a porous barrier, how much time will it take for the same amount

alisha [2963]

Answer:

The correct response is;

2. takes less than 15 seconds

Explanation:

Graham's law of effusion conveys that

$\frac{Rate 1}{Rate2} =\sqrt{\frac{Molar Mass2}{MolarMass 1} }$

Thus, knowing that the molar mass of neon gas = 20.18 g/mol

And, the molar mass of nitrogen gas = 28.014 g/mol

We deduce

= 0.72

$\frac{Rate 1}{15} =\sqrt{\frac{20.18}{28.014} }$ So, the effusion rate of nitrogen gas will be 0.72*15 = 12.73 seconds, which is indeed less than 15 seconds.

5 0

8 days ago

Using this information together with the standard enthalpies of formation of O2(g), CO2(g), and H2O(l) from Appendix C, calculat

Tems11 [2777]

The question is not fully stated; here is the full version:

Using this data alongside the standard enthalpies of formation for $O_2(g)$ , $CO_2(g)$ , and $H_2O(l)$ found in Appendix C, determine the standard enthalpy of formation for acetone.

The complete combustion of 1 mole of acetone $(C_3H_6O)$ releases 1790 kJ:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l);\Delta H^o=-1790kJ$

Answer: The standard enthalpy of formation for $CO_2(g)$ is calculated to be -247.9 kJ/mol

Explanation:

Enthalpy change represents the variation in enthalpy for all products and reactants based on their respective mole counts. This is denoted as $\Delta H^o$

The enthalpy change calculation for a chemical reaction follows this equation:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

Concerning the chemical reaction in question:

$C_3H_6O(l)+4O_2(g)\rightarrow 3CO_2(g)+3H_2O(l)$

The equation reflecting the enthalpy change for this reaction is:

$\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(3\times \Delta H^o_f_{(H_2O(l))})]-[(1\times \Delta H^o_f_{(C_3H_6O(l))})+(4\times \Delta H^o_f_{(O_2(g))})]$

Provided data includes:

$\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=-1790kJ$

Substituting values from the equation gives us:

$-1790=[(3\times {(-393.5)})+(3\times (-285.8))]-[(1\times \Delta H^o_f_{(C_3H_6O(g))})+(4\times (0))]\\\\\Delta H^o_f_{(C_3H_6O(g))}=-247.9kJ/mol$

Thus, the enthalpy of formation of $C_3H_6O(g)$ computes to -247.9 kJ/mol.

4 0

1 month ago

Assuming that only the listed gases are present, what would be the mole fraction of oxygen gas be for each of the following situ

eduard [2782]

Mole fraction of oxygen gas: 0.381

Additional clarification

Given:

2.31 atm Oxygen

3.75 atm Hydrogen

Required:

Mole fraction of Oxygen

Calculation:

According to Dalton’s Law of partial pressures

P tot = P₁ + P₂ +.. + Pₙ

Substituting values:

P tot = P O₂ + P H₂

P tot = 2.31 atm + 3.75 atm

P tot = 6.06 atm

Mole fraction of O₂ (X O₂):

P O₂ = X O₂ x P tot

X O₂ = P O₂ / P tot

X O₂ = 2.31 / 6.06

X O₂ = 0.381

6 0

1 month ago

A magnesium ion, Mg2+, with a charge of 3.2×10−19C and an oxide ion, O2−, with a charge of −3.2×10−19C, are separated by a dista

eduard [2782]

Details:

The equation to calculate work done is defined as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

where, k = proportionality constant = $8.99 \times 10^{9} Jm/C^{2}$

$q_{1} = charge of Mg^{2+} = 3.2 \times 10^{-19} C$

$q_{2} = charge of O_{2-} = -3.2 \times 10^{-19} C$

d = separation distance = 0.45 nm = $0.45 \times 10^{-9} m$

Now we will insert the given values into the formula above to compute the work done as follows.

W = $-k \frac{q_{1}q_{2}}{d}$

= $\frac{-[8.99 \times 10^{9} Jm/C^{2} \times 3.2 \times 10^{-19} C \times -3.2 \times 10^{-19} C]}{0.25 \times 10^{-9} m}$

= $3.68 \times 10^{-18} J$

Thus, we can conclude that the work needed to increase the distance between the two ions to infinity is $3.68 \times 10^{-18} J$ .

7 0

1 month ago

Solid aluminum hydroxide reacts with a solution of hydrobromic acid. write a balanced molecular equation and a balanced net ioni

Tems11 [2777]

<span>To determine the balanced chemical equation, it is necessary to identify the symbols and charges for each element and then balance the equation. This results in: Al(OH)3(aq) + 3HBr(aq) ---> AlBr3(aq) + 3H2O(l) The net ionic equation involves summing the charges and breaking them down into fundamental parts. Al3+ + 3OH- + 3H+ + 3Br- --> Al3+ + 3Br + H3O+ After eliminating aluminum and bromine, we have: 3OH-(aq) + 3H+(aq) --> 3H2O(l)</span>

4 0

14 days ago

Read 2 more answers