Balance the equation by typing in the correct coefficients into the blank spaces in front of each formula. Type 1 if

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

Plz answer thx<br><br><br> How many moles are in 2.5 x 10^18 grams of water?

alex41 [277]

Answer:

$1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

Explanation:

Mass of water = $m = 2.5\times 10^{18} g$

Molar mass of water = M = 18 g/mol

Moles = n = $\frac{m}{M}$

$n=\frac{2.5\times 10^{18} g}{18 g/mol}=1.39\times 10^{17} moles$

So, there are $1.39\times 10^{17}$ moles of water in $2.5\times 10^{18} g$ of water.

3 0

2 years ago

Spell out the full name of the compound.

Talja [164]

Answer:

2–methyl–3–propylheptane.

Explanation:

The name of the compound can be obtained as follow:

1. Identify the functional group of the compound.

2. Determine the name of the compound by determining the longest continuous carbon chain.

3. Identify the substituent group attached.

4. Locate the position of the substituent group by giving it the lowest possible count. Where there are two or more different substitutent group attached to the compound, name them alphabetically.

5. Name the compound by combining

the above steps.

Thus, we can name the compound as follow:

1. The compound has only single bond. Therefore, the compound is an alkane.

2. The longest continuous carbon chain is 7. Therefore, the parent name of the compound is heptane.

3. The substitutent groups attached are:

Propyl i.e –C3H7

Methyl i.e –CH3

4. The substituent groups are located at:

Propyl i.e –C3H7 is located at carbon 3.

Methyl i.e –CH3 is located at carbon 2.

5. The name of the compound is:

2–methyl–3–propylheptane.

6 0

2 years ago

What is the cost to drive to Los Angeles from San Francisco a distance of 405 miles if the cost of gasoline is $2.90 a gallon an

choli [55]

Answer: $46.98

Explanation:

The distance to drive to Los Angeles from San Francisco is 405 miles. Since the car get 25 miles per gallon of gas, for 405 miles, the car will use:

= 405/25 = 16.2 gallons

Since the cost of gasoline is $2.90 a gallon and the car will use 16.2 gallons. The cost will be:

= $2.90 × 16.2 gallons

= $46.98

7 0

2 years ago

Using the balanced equation N2+O2=2NO, how many grams of NO can be produced when 25.0 grams of N react?

Oxana [17]

Answer:

$53.55gNO$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it is possible for us to calculate the produced grams of nitrogen monoxide by starting with 25.0 g of nitrogen via their 1:2 mole ratio and the molar masses of 30.1 g/mol and 28.02 g/mol, respectively and by some stoichiometry:

$=25.0gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNO}{1molN_2}*\frac{30.01 gNO}{1molNO}\\\\=53.55gNO$

Best regards!

8 0

2 years ago