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sashaice [31]
2 years ago
10

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

ur answer, take the negative log and enter that (so it's like you're taking the pKsp)
Chemistry
1 answer:
Snezhnost [94]2 years ago
4 0

<u>Answer:</u> The solubility product of silver (I) phosphate is 9.57\times 10^{-10}

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)  

                            3s                  s

The expression of K_{sp} for above equation follows:

K_{sp}=(3s)^3\times s

We are given:  

s=2.44\times 10^{-3}M

Putting values in above expression, we get:

K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}

Hence, the solubility product of silver (I) phosphate is 9.57\times 10^{-10}

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. The net ionic equation for the reaction that occurs during the titration of nitrous acid with sodium hydroxide is (a) HNO2 Na
hjlf

Answer:

H^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Only the species which are present in aqueous state dissociate.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

(a)

The balanced molecular equation will be,

HNO_2_{(aq)}+NaOH_{(aq)}\rightarrow H_2O_{(l)}+NaNO_2_{(aq)}

The complete ionic equation in separated aqueous solution will be,

H^+_{(aq)}+NO_2^-_{(aq)}+Na^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}+Na^+_{(aq)}+NO_2^-_{(aq)}

In this equation the species present are, Na^+\text{ and }NO_2^- are the spectator ions.

Hence, the net ionic equation contains specie is  

H^+_{(aq)}+OH^{-}_{(aq)}\rightarrow H_2O_{(l)}

6 0
2 years ago
Show your calculation by uploading a picture. Calculate the molar mass of ammonia, NH3
Cloud [144]

Answer:

17.04 g/mol

Explanation:

Molar Mass of NH₃

we know that

Nitrogen has 14.01 gram/mol

And Hydrogen has 1.01 gram/mol

but we have 3 Hydrogens So we multiply

1.01 by 3 i.e., 3.03

Now, add

14.01

+<u> </u><u>3</u><u>.</u><u>0</u><u>3</u>

17.04

So, The molar mass of ammonia, NH₃ is

17.04 g/mol

<u>-TheUnknown</u><u>Scientist</u>

5 0
3 years ago
How many times do u have to run back and forth across a football field to be meter
Taya2010 [7]
You would have to run 100 kilometers across it once. Then you will have to run 10 meters across it because a football field is 110 meters according to the NFL’s website
7 0
2 years ago
(i) Based on the graph, determine the order of the decomposition reaction of cyclobutane at 1270 K. Justify your answer.
Leni [432]

Answer:

(c)(i) The order of the reaction based on the graph provided is first order.

(ii) 99% of the cyclobutane would have decomposed in 53.15 milliseconds.

d) Rate = K [Cl₂]

K = rate constant

The justification is presented in the Explanation provided below.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Explanation:

To investigate the order of a reaction, a method of trial and error is usually employed as the general equations for the amount of reactant left for various orders are known.

So, the behaviour of the plot of maybe the concentration of reactant with time, or the plot of the natural logarithm of the concentration of reactant with time.

The graph given is evidently an exponential function. It is a graph of the concentration of cyclobutane declining exponentially with time. This aligns with the gemeral expression of the concentration of reactants for a first order reaction.

C(t) = C₀ e⁻ᵏᵗ

where C(t) = concentration of the reactant at any time

C₀ = Initial concentration of cyclobutane = 1.60 mol/L

k = rate constant

The rate constant for a first order reaction is given

k = (In 2)/T

where T = half life of the reaction. It is the time taken for the concentration of the reactant to fall to half of its initial concentration.

From the graph, when the concentration of reactant reaches half of its initial concentration, that is, when C(t) = 0.80 mol/L, time = 8.0 milliseconds = 0.008 s

k = (In 2)/0.008 = (0.693/0.008) = 86.64 /s

(ii) Calculate the time, in milliseconds, that it would take for 99 percent of the original cyclobutane at 1270 K to decompose

C(t) = C₀ e⁻ᵏᵗ

when 99% of the cyclobutane has decomposed, there's only 1% left

C(t) = 0.01C₀

k = 86.64 /s

t = ?

0.01C₀ = C₀ e⁻ᵏᵗ

e⁻ᵏᵗ = 0.01

In e⁻ᵏᵗ = In 0.01 = -4.605

-kt = -4.605

t = (4.605/k) = (4.605/86.64) = 0.05315 s = 53.15 milliseconds.

d) The reaction mechanism for the reaction of cyclopentane and chlorine gas is given as

Cl₂ → 2Cl (slow)

Cl + C₅H₁₀ → HCl + C₅H₉ (fast)

C₅H₉ + Cl → C₅H₉Cl (fast)

The rate law for a reaction is obtained from the slow step amongst the the elementary reactions or reaction mechanism for the reaction. After writing the rate law from the slow step, any intermediates that appear in the rate law is then substituted for, using the other reaction steps.

For This reaction, the slow step is the first elementary reaction where Chlorine gas dissociates into 2 Chlorine atoms. Hence, the rate law is

Rate = K [Cl₂]

K = rate constant

Since, no intermediates appear in this rate law, no further simplification is necessary.

The obtained rate law indicates that the reaction is first order with respect to the concentration of the Chlorine gas and zero order with respect to cyclopentane.

e) A catalyst is a substance that alters the rate of a reaction without participating or being used up in the reaction.

Cl₂ is one of the reactants in the reaction, hence, it participates actively and is used up in the process of the reaction, hence, it cannot be termed as a catalyst for the reaction.

So, this shows why the student's claim is false.

Hope this Helps!!!

6 0
3 years ago
How many O2 molecules react with 47 CH4 molecules according to the preceding equation CH4 + O2 → CO2 + H2O
natka813 [3]
I think 93.71 mol.
You need to use stochio

8 0
3 years ago
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