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luda_lava [24]
3 years ago
8

When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II)

sulfate PbSO4 is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode. Suppose a current of 17.0A is fed into a car battery for 15.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Be sure your answer has a unit symbol and the correct number of significant digits.
Chemistry
1 answer:
zysi [14]3 years ago
7 0

Answer:

0.273504 grams

Explanation:

Charge Deposited = Current ×Time

q=It

Given: I= 17.0A, t= 15.0 seconds

Q= 17×15= 255 C

Since, 1 mole contains 96500C of charge deposition

Therefore, number of moles = 255/96500 = 0.00264 moles

Moreover, since Pb changes to Pb2+,

Hence number of moles of Pb deposited = 0.00264/2

= 0.00132

Also, Molar mass of Lead = 207.2 gm/mol

Therefore, the mass of lead deposited = 0.00132×207.2 = 0.273504 grams

.

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A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm
Ulleksa [173]

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100

Percent error = \frac{0.1}{63.2}\times 100

Percent error = 0.00158\times 100

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100

Percent error = \frac{0.2}{63.2}\times 100

Percent error = 0.00316\times 100

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= \frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100

Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100

Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100

Percent error = \frac{(0.5)}{63.2}\times 100

Percent error = 0.00791\times 100

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%




7 0
3 years ago
What 3 gases does the flame test check for and what should you expect to see/hear?
Kazeer [188]
What of the gasses is N and the flame changes it's color from Orange to blue
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3 years ago
1. Addition of which of the following will increase the solubility of CaCO3 in water? Consider the equilibrium process:
kolbaska11 [484]

Answer: HCl

Explanation:

calcium carbonate dissolves in HCl acid producing CO 2 gas. It will not dissolve in pure water. The Ksp for calcium carbonate in water is 3.4 x 10-9 moldm-3 which is very low. What takes place here is actually a chemical reaction:

CaCO 3 (s) + 2HCl(aq) → CaCl 2 (aq) + H 2CO 3(aq)

This reaction accounts for the solubility of the Calcium carbonate in HCl and not in pure water.

6 0
3 years ago
Un anillo, de masa 90 gramos, contiene 59,1% de oro. ¿Cuál es el valor del anillo si cada mol de oro vale S/.1800?. Considere el
LekaFEV [45]

Answer:

S/.486 es el valor del anillo

Explanation:

Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.

Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:

90g × 59.1% = 53.19g Oro en el anillo

Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:

53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.

Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:

0.27mol × (S/.1800 / 1mol oro) =

<h3>S/.486 es el valor del anillo</h3>
8 0
2 years ago
The expected o-c-h angle in this molecule is degrees. the expected hybridization at the central carbon is
WINSTONCH [101]

Each neutral carbon atom contains four valence electrons and may form up to four electron domains. Possible hybridizations include

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Molecules of each of the three hybridization demonstrate spatial configurations that would maximizes the separation between the electron domains.

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Bond angles are characteristic of the spatial configuration of electron domains and identifies the hybridization of the central carbon atom.

Note that each hydrogen atom contains only one valence electron and would form only single bonds. It takes two valence electrons for oxygen atoms to achieve an octet such that each oxygen form only two bonds at a single time. Therefore given the fact that the carbon is bonded to both hydrogen and oxygen, only the following hybridizations are possible

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3 0
3 years ago
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