All categories

3 years ago

describe the differences of the kinetic energy of a cold water solution to that of a hot water solution

Chemistry

2 answers:

hjlf3 years ago

8 0

Answer:

Here are the differences between the hot and cold water. The particles in the hot water move faster than the particles in the cold water.

Explanation: I hope this helped you!!!!!!!:)

Natasha_Volkova [10]3 years ago

4 0

The kinetic energy is in relation to the temperature. In cold water, the molecules in the water moves slower and gives off less heat. Which also means it has a lower temperature and lower kinetic energy. Hot water is the opposite. The molecules move faster and have a higher kinetic energy. It has a higher temperature too.

You might be interested in

Taxol is a potent chemotherapeutic agent (isolated from the Pacific Yew tree) which is especially effective against ovarian canc

butalik [34]

Answer:

Amine

Explanation:

The functional groups contained in Taxol are :

Ketone , Ester, Amide and Alcohol

while the functional group that is not contained in the Taxol is Amine

Taxol is a very potent anti-cancer chemotherapeutic, and it is also groped into a class called Taxanes and this makes it effective in the treatment of breast and ovarian cancer.

8 0

3 years ago

Which statement regarding liquids is true? 1.can be converted to solids by adding energy in the form of heat 2.occupy a definite

GuDViN [60]

Answer:

4.occupy a definite volume while taking the shape of their container.

Explanation:

Since the particles in a liquid are quite close together (not as close as a solid) they have a definete volume, and the particles are still able to move around as they wish they have no definite shape and take upon the shape of the container they are placed in.

8 0

3 years ago

How many grams of Cl are in 465g of CaCl2

Rama09 [41]

2 ways to do this
a. find %Cl in CaCl2
2 x 35.45g/mole = 70.9g Cl
70.9g Cl / 110.9g/mole CaCl2 = 63.93% Cl in CaCl2
0.6963 x 145g = 92.7g = mass Cl

b. determine moles CaCl2 present then mass Cl
145g / 110.9g/mole = 1.31moles CaCl2 present
2moles Cl / 1mole CaCl2 x 1.31moles = 2.62moles Cl
2.62moles Cl x 35.45g/mole = 92.7g Cl

6 0

3 years ago

Read 2 more answers

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

4 years ago

How many atoms of magnesium metal are needed to produce 3.87 moles of magnesium oxide

Ket [755]

Answer:

48.42g of Mg

Explanation:

4 0

3 years ago

Read 2 more answers