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3 years ago

What is the concentration of each ion in 15.00 ml of a 7.85 X 10^-6 M solution of Tc3 (PO4)7?

Chemistry

1 answer:

Daniel [21]3 years ago

5 0

Answer:

Tc^(+7) - 2.35*10^-5M

PO4(^-3) - 5.50*10^-5M

Explanation:

Tc3(PO4)7 <---> $3Tc^{7+} + 7PO_{4}^{3-}$

1 mol 3 mol 7 mol

7.85 X 10^-6 M 3*7.85*10^-6M 7*7.85*10^-6M

2.35*10^-5M 5.50*10^-5M

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Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce

levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M $Ca^{2+}(aq)$ and 0.0390 M $Ag^{+}(aq)$ . What will be the concentration of $Ca^{2+}$ (aq) when $Ag_{2}SO_{4}(s)$ begins to precipitate? What percentage of the $Ca^{2+}(aq)$ can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

$Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}$

$[Ag^{+}]$ = 0.0390 M

When $Ag_{2}SO_{4}$ precipitates then expression for $K_{sp}$ will be as follows.

$K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]$

$1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]$

$[SO^{2-}_{4}]$ = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

$CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}$

$K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

$4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788$

$[Ca^{2+}]$ = 0.00625 M

Now, we will calculate the percentage of $Ca^{2+}$ remaining in the solution as follows.

$\frac{0.00625}{0.05} \times 100$

= 12.5%

And, the percentage of $Ca^{2+}$ that can be separated is as follows.

100 - 12.5

= 87.5%

Thus, we can conclude that 87.5% will be the concentration of $Ca^{2+}(aq)$ when $Ag_{2}SO_{4}(s)$ begins to precipitate.

4 0

3 years ago

Wind blowing sediment against rock or land area or sand blasting is called

artcher [175]

The correct answer is - deflation.

The process of deflation can be caused by the winds. It is an erosive process in which the main role has the wind that is carrying lot of sediment in the shape of very small particles with it.

Through this process, the winds manage to erode large areas, especially in the drier places where the vegetation is very sparsely distributed. By this type of erosion, the winds manage to make lot of hollows that can range significantly in size. The hollows made by the deflation can be anywhere from few cm deep and several meters long, up to several km long and 50-60 meters of depth.

This is the process that is responsible for the creation of most of the oasis in the largest desert in the world, Sahara, some even being lowered enough to be under the sea level.

7 0

3 years ago

32 gm of O2 to mole of O2

lana [24]

Molar mass of O2:-

$\\ \rm\longmapsto 2(16u)=32g/mol$

Now

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto No\:of\;moles=\dfrac{32}{32}$

$\\ \rm\longmapsto No\:of\;moles=1mol$

5 0

3 years ago

A certain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k . what is the ratio of f at the h

lapo4ka [179]

Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:

K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency..

Then you can find the ration between K's at two different temperatures as:

K1 = A * e ^ (-Ea / RT1)

K2 = A* e ^(-Ea / RT2)

=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }

=> K1 / K2 = e ^ {(-Ea/ R ) *( 1 / T1 - 1 T2) }

=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k )* ( 1 / 505K - 1/ 485K) }

=> K1 / K2 = e ^ (2.0134494) ≈ 7.5

Answer: 7.5

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3 years ago

A sample of gas occupies 10.0 l at 100.0 torr and 27.0

disa [49]

The pressure of a sample of a gas if the temperature is changed to 127 c while the volume remains constant is calculated using gay lussac law formula

that is P1/T1 = P2/V2
P1 = 100 torr
T1 = 27+273 = 300 k
T2 =127 +273 =400 k
P2=?

by making P2 the subject of the formula
P2=T2P1/T1

=100 x 400/300 = 133.3 torr

5 0

4 years ago