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2 years ago

WORTH 20 POINTS HELP PLEASE

Chemistry

1 answer:

Damm [24]2 years ago

5 0

Answer:

The answers to your questions are below

Explanation:

Physical change is when matter changes its form but is still the same substance.

Chemical change is when matter changes its composition.

iron and oxygen from rust It's a chemical change because iron and oxygen form a new compound different from the reactants.

iron is denser brightly when ignited . This is a chemical property because iron reacts with oxygen.

oil and water do not mix it's a physical change, oil and water are in touch but they do not react.

mercury melts at -39°C it's a physical property because mercury changes from solid to liquid but it still is mercury.

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What belief do Deists hold about the universe?

liubo4ka [24]

"b. God had created a mechanistic universe that could only be understood through the Bible" is the best option since Deists believe God treats the universe in a "hands off" way.

8 0

2 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

3 years ago

B. The following reaction takes place in a basic solution. (7 points)

Natali5045456 [20]

Answer: The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

$MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)$

To balance the given redox reaction in basic medium, there are few steps to be followed:

Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$

Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$ ( × 3)

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$ ( × 2)

The half-reactions now become:

Oxidation half-reaction: $3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-$

Reduction half-reaction: $2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-$

Add the equations and simplify to get a balanced equation

Overall redox reaction: $3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-$

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

3 0

2 years ago

How many electrons does nitrogen (N) need to gain to have a stable outer electron shell

earnstyle [38]

It need "3 electrons" to have a stable electronic configuration.

'cause it has 5 electrons in it's outer shell & every atom needs 8 electrons. So, it requires 3 more!

Hope this helps!

7 0

2 years ago

Read 2 more answers

6. A. If 4.50 mols of ethane, C2H6, undergoes combustion according to the unbalanced equation

uranmaximum [27]

Answer:

for one mole of C2H6 there are 7/2 mole of O2 required. so for4. 50 moles you require 4.50 x 7/2 = 15.75 moles of O2.

Explanation:

i hope it's helpful

6 0

2 years ago