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lora16 [44]
3 years ago
5

The accepted value for the molar volume of a gas is 22.4 L. Sarah's experimental data indicated that a mole of a gas had a volum

e of 19.6L. What was her percent error?
Chemistry
1 answer:
Amanda [17]3 years ago
3 0

Answer:

Percent error = 12.5%

Explanation:

In a measurement you can find percent error following the formula:

Percent error = |Measured value - Accepted Value| / Acepted value * 100

Based on the data of the problem, accepted value is 22.4L and the measured Value (Value of Sara) was 19.6L.

Replacing:

Percent error = |Measured value - Accepted Value| / Acepted value * 100

Percent error = |19.6L - 22.4L| / 22.4L * 100

Percent error = |-2.8L| / 22.4L * 100

Percent error = 2.8L / 22.4L * 100

Percent error = 12.5%

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crimeas [40]

Answer:

cytoplasm is A, Dna is C, and nucleas is B

Explanation:

3 0
3 years ago
Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v​
bogdanovich [222]

Answer:

Depending on the E^\circ value of \rm Ag^{+} + e^{-} \to Ag\; (s), the cell potential would be:

  • 0.55\; \rm V, using data from this particular question; or
  • approximately 0.46\; \rm V, using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

  • The conversion between \rm Ag\; (s) and \rm Ag^{+} ions, \rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s), and
  • The conversion between \rm Cu\; (s) and \rm Cu^{2+} ions, \rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s).

Note that the standard reduction potential of \rm Ag^{+} ions to \rm Ag\; (s) is higher than that of \rm Cu^{2+} ions to \rm Cu\; (s). Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if \rm Ag^{+} ions are reduced while \rm Cu\; (s) is oxidized.

Therefore:

  • The reduction reaction at the cathode will be: \rm Ag^{+} + e^{-} \to Ag\; (s). The standard cell potential of this reaction (according to this question) is E(\text{cathode}) = 0.89\; \rm V. According to the 2012 CRC handbook, that value will be approximately 0.79\; \rm V.
  • The oxidation at the anode will be: \rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}. According to this question, this reaction in the opposite direction (\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)) has an electrode potential of 0.34\; \rm V. When that reaction is inverted, the electrode potential will also be inverted. Therefore, E(\text{anode}) = -0.34\; \rm V.

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}.

Using data from the 1985 and 2012 CRC Handbook:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}.

5 0
3 years ago
Please help! Thanks
GarryVolchara [31]
I believe the third choice is correct.

This can be proven by the fact that to find the molar mass of a compound, you simply add the molar masses of all the atoms within the compound

Hope this helps
5 0
3 years ago
Read 2 more answers
General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.
Helen [10]

Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

  • Q: heat absorbed
  • C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
  • ΔT: temperature change (2.29 °C)

Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ

According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

We will use the conversion factor 1 Cal = 4.186 kJ.

86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

Step 3: Calculate the number of Cal per gram of candy

20.6 Cal/6.70 g = 3.07 Cal/g

3 0
3 years ago
How can magnetic force be exerted on objects?
just olya [345]
I think the answer is : 1
8 0
2 years ago
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