Calculate the pH and fraction of dissociation (

A standard 10.00 g mass is weighed on an analytical balance 100 times. The average and standard deviation obtained gives 10.12 ±

Mekhanik [1.2K]

Answer:

There was an improvement in accuracy. There was no change in precision.

Explanation:

The average mass after recalibration is closer to the mass of the standard, so the recalibration improved the accuracy (the measurement is closer to an accepted 'true' value).

The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.

5 0

3 years ago

Which of the following information is necessary to describe an object's motion?

Zepler [3.9K]

Answer: The correct option is B.

Explanation: To describe the motion of an object, we use the equations of motion.

$v=u+at$
$s=ut+\frac{1}{2}at^2$
$2as=v^2-u^2$

From the above equations, we require position, speed and direction through which we an calculate the displacement, velocity and acceleration.

To calculate the complete motion of an object, we require all the three factors.

Hence, the correct option is B.

8 0

2 years ago

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Chemical energy is a form of A. kinetic energy only. B. both potential and kinetic energy. C. neither potential nor kinetic ener

svetlana [45]

wrong. If its for the study island its potential energy only.

6 0

3 years ago

A chemical reaction that is expected to form 325.0 gof product only forms 123.8 g of product. What is the percent yield of this

Tasya [4]

Answer:

$\boxed {\boxed {\sf A. \ 38.1 \%}}$

Explanation:

Percent yield is the ratio of the amount actually produced to how much could theoretically be produced. It is found using this formula:

$\% \ yield = \frac{actual \ yield}{theoretical \ yield} *100$

For this reaction, the theoretical or expected yield is 325.0 grams. The actual yield is 123.8 grams.

$\% \ yield = \frac{ 123.8 \ g }{325.0 \ g }*100$

Divide.

$\% \ yield = 0.380923077 *100$

$\% \ yield = 38.0923077$

Round to the nearest hundredth. The 9 in the hundredth place tells us to round the 0 to a 1 .

$\% \ yield \approx 38.1$

The percent yield is about 38.1%

5 0

2 years ago

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Write the net cell equation for this electrochemical cell. Phases are optional. Do not include the concentrations. Co ( s ) ∣ ∣

Tatiana [17]

Answer: The overall equation will be $Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

Explanation:

The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Anode : $Co(s)\rightarrow Co^{2+}(aq)+2e^{-}$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Cathode : $Ag^+(aq)+e^{-1}\rightarrow Ag(s)$ $\times 2$

The number of electrons lost must be equal to the number of electrons gained , thus overall equation will be :

$Co(s)+2Ag^+(aq)\rightarrow Co^{2+}(aq)+Ag(s)$

5 0

2 years ago