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Tatiana [17]
3 years ago
9

Which of the following reasons (specific to this lab) would have caused the percent yield of calcium carbonate to be less than 1

00%? Highlight all that apply (there may be more than one)
Some of the sodium carbonate never fully dissolved in the water and did not transfer when mixed with calcium acetate

The glassware used to mix the reactant solutions were not properly dried before use

Some of the calcium carbonate remained stuck to the side of the glassware after filtering

The product, calcium carbonate, was not fully dried out before weighing

Some of the calcium carbonate spilled when measuring its mass
Chemistry
1 answer:
mote1985 [20]3 years ago
8 0

Answer:

Some of the calcium carbonate remained stuck to the side of the glassware after filtering

Some of the calcium carbonate spilled when measuring its mass

Explanation:

The percent yield is obtained from;

Actual yield/Theoretical yield * 100

Usually, the actual yield is less than the theoretical yield thereby making the percent yield less than 100%.

In the case of this particular reaction; when some calcium carbonate remains stuck on the glassware or some calcium carbonate is spilled during weighing, the actual yield of product obtained from the reaction becomes less than the theoretical or calculated yield hence the percent yield becomes less than 100%.

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Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration
gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}

        K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}

Let us assume that the solubility be "s". And, the reaction equation is as follows.

        AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

     9.84 \times 10^{-21} = s \times s

             s = 9.92 \times 10^{-11}

Also,     Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}

                2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}

                            s = 7.14 \times 10^{-7}

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the K_{sp} expression as follows.

         K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}

          2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}

       2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}

       [PO^{3-}_{4}]^{2} = 4.88 \times 10^{-24}

                             = 2.21 \times 10^{-12} M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

             AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

           K_{sp} = [Al^{3+}][PO^{3-}_{4}]

       9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}

                [Al^{3+}] = 4.45 \times 10^{-9} M

Thus, we can conclude that concentration of aluminium will be 4.45 \times 10^{-9} M when calcium begins to precipitate.

7 0
3 years ago
The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,
Masja [62]

<u>Answer:</u> The concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

<u>Explanation:</u>

The given cell is:

Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V ( × 3)

<u>Reduction half reaction:</u> Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V ( × 2)

<u>Net reaction:</u> 3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=1.50-0=1.50V

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}

where,

E_{cell} = electrode potential of the cell = 1.23 V

E^o_{cell} = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

[Au^{3+}]=?M

[H^{+}]=1.0M

Putting values in above equation, we get:

1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})

[Au^{3+}]=1.87\times 10^{-14}M

Hence, the concentration of Au^{3+} in the solution is 1.87\times 10^{-14}M

7 0
3 years ago
An over-the-counter multi-vitamin has 350 mg tablets that are 20% calcium by mass. How many tablets are required for you to take
Marina86 [1]
.2x350mg = 70mg which converts to .070g.
.35g/.070g = 5 tablets.
7 0
3 years ago
Read 2 more answers
Velocity is the slope of the acceleration vs. time graph.<br>O<br>A. True<br>False​
Maurinko [17]

Answer:

False

Explanation:

The slope of a velocity-time graph gives acceleration. Acceleration can be defined as the change in velocity with time.

A slope denotes the gradient of line. It takes into consideration the changes on both y and x axis. The ratio of the changes gives the slope.

On a velocity-time graph, the y-axis is the velocity and the x-axis is time. The change in velocity with time gives acceleration.

The slope of an acceleration

-time graph is not velocity.

7 0
4 years ago
The atomic weight of Ga is 69.72 amu. There are only two naturally occurring isotopes of gallium:
irga5000 [103]

Answer:

71 Ga has a naturally abundance of 36%

Explanation:

Step 1: Given data

Gallium has 2 naturally occurring isotopes: this means the abundance of the 2 isotopes together is 100 %. The atomic weight of Ga is 69.72 amu. This is the average of all the isotopes.

Since the average mass of 69.72 is closer to the mass of 69 Ga, this means 69 Ga will be more present than 71 Ga

Percentage 69 Ga> Percentage 71 Ga

<u>Step 2:</u> Calculate the abundance %

⇒Percentage of 71 Ga = X %

⇒Percentage of 69 Ga = 100 % - X %

The mass balance equation will be:

100*69.72 = x * 71 + (100 - x)*69

6972 = 71x + 6900 -69x

72 = 2x

x = 36 %

71 Ga has a naturally abundance of 36%

69 Ga has a naturally abundance of 64%

3 0
4 years ago
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