The number 0.00003007 expressed in exponential notation is

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

2 years ago

Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cooki

Leni [432]

Answer:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = $\frac{936}{8}$ = 117

By substituting the value of (y+z) in equation- (1) we get,

x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

y = $\frac{535-(6\times 32)}{7}$ = 49

By substituting the value of y into equation- (2) we get,

z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

6 0

3 years ago

What features make a question scientific, and why are some questions beyond the boundaries of scientific investigation?

jeka57 [31]

The scientific method

8 0

3 years ago

theres this really cute boy in my science class and i asked him if he wanted to be my lab partner and he said yes he doesn't tal

natita [175]

Make small talk and become his friend get to know him if he opens up he trust you and then you take it from there.

3 0

3 years ago

Read 2 more answers

How many moles of carbon are in 3.7 moles of C 8 H11 NO2

Dafna1 [17]

In every molecule of $C_{8} H_{11}NO_2$ there is 8 atoms of Carbon.

IF we have 3.7 moles of $C_{8} H_{11}NO_2$ to find the number of moles of Carbon, just multiply by 8

3.7 * 8 = 29.6 mol Carbon

6 0

2 years ago