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3 years ago

5

A solution is a special mixture where you cannot see the individual substances that are mixed. You look at two bowls of ice crea

m. One bowl is plain vanilla ice cream and the other is vanilla with chocolate chips. Which statement accurately describes the two bowls of ice cream?
A) Both ice creams bowls are solutions.
B) Neither are mixtures because they are solid.
C) Only the chocolate chip ice cream is a mixture.
D) Both are mixtures but only the vanilla is a solution.

1 answer:

Nutka1998 [239]3 years ago

3 0

Wouldn't be a then not c if both were solutions.

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Given the following reaction: \ce{Cu + 2AgNO3 -> 2Ag + Cu(NO3)2}Cu+2AgNO3 2Ag+Cu(NOX 3 )X 2 How many moles of \ce{Ag}

Alexandra [31]

Answer:

0.252 mol

Explanation:

<em>Given the following reaction: </em>

<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>

<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 is available in excess.</em>

First, we write the balanced equation.

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

We can establish the following relations.

The molar mass of Cu is 63.55 g/mol.
The molar ratio of Cu to Ag is 1:1.

The moles of Ag produced from 16.0 g of Cu are:

$16.0gCu.\frac{1molCu}{63.55gCu} .\frac{1molAg}{1molCu} =0.252 molAg$

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3 years ago

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What is the salt that is produced when calcium hydroxide (Ca(OH)2) reacts with sulfuric acid (H2SO4)? CaSH2Ca H2O CaSO4

Wittaler [7]

H2SO4+CA[OH]2=CASO4+2H2O

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3 years ago

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Which of the following is an example of a change of state?

lord [1]

Answer:

Option B

Explanation:

For Option A the state is not changing but just has different look now.
For Option B the state is changing from gas to liquid drops due to cold glass
For Option C it hasn't changed state as both corn and flower is solid
For Option D paper and ash are both solids

Therefore our answer must be Option B

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2 years ago

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%

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3 years ago

The earth rotates on an imaginary line called an?

Pavel [41]

Answer: An axis.

Explanation:

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4 years ago

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