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seropon [69]
4 years ago
9

Given the following reaction: \ce{Cu + 2AgNO3 -> 2Ag + Cu(NO3)2}Cu+2AgNO3 ​ ​ 2Ag+Cu(NOX 3 ​ )X 2 ​ How many moles of \ce{Ag}

AgA, g will be produced from 16.0 \text{ g}16.0 g16, point, 0, start text, space, g, end text of \ce{Cu}CuC, u, assuming \ce{AgNO3}AgNO3 ​ is available in excess
Chemistry
2 answers:
Alexandra [31]4 years ago
6 0

Answer:

0.252 mol

Explanation:

<em>Given the following reaction: </em>

<em>Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂</em>

<em>How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 ​ is available in excess.</em>

First, we write the balanced equation.

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

We can establish the following relations.

  • The molar mass of Cu is 63.55 g/mol.
  • The molar ratio of Cu to Ag is 1:1.

The moles of Ag produced from 16.0 g of Cu are:

16.0gCu.\frac{1molCu}{63.55gCu} .\frac{1molAg}{1molCu} =0.252 molAg

sweet [91]4 years ago
4 0

Answer:

n_{Ag} =0.504molAg

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Cu+2AgNO_3-->Cu(NO_3)_2+2Ag

As the silver nitrate is in excess, the yielded moles of silver are computed from the 16g of copper as shown below, including the corresponding stoichiometric 1 to 2 relationship between copper and silver respectively:

n_{Ag} =16.0gCu*\frac{1molCu}{63.5gCu}*\frac{2molAg}{1molCu}\\n_{Ag} =0.504molAg

Best regards.

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babunello [35]

Answer:

91 millilitres

Explanation:

Recommended application = 65mg / Kg

This means 65 mg of dicyclanil per kg (1 kg of body mass).

Concentration = 50 mg / mL

How many millilitres required to treat 70kg adult?

If 65mg = 1 kg

x = 70 mg

x = 70 * 65 = 4550 mg

Concentration = Mass / Volume

50 mg/mL = 4550 / volume

volume = 4550 / 50 = 91 mL

8 0
3 years ago
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150
Effectus [21]

<u>Given:</u>

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

<u>To determine:</u>

The cell potential, Ecell

<u>Explanation:</u>

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻                E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s)              E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

8 0
3 years ago
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1. How many calcium atoms are in 8CaOH?
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Answer:

8

Explanation:

the number before the abbrviation for calcium (Ca) shows how many calcium there is.

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3 years ago
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A 21.8 g sample of ethanol (C2H5OH) is burned in a bomb calorimeter, according to the following reaction. If the temperature ris
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<u>Answer:</u> The heat capacity of calorimeter is 15.66J/^oC

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of ethanol = 21.8 g

Molar mass of ethanol = 46.07 g/mol

Putting values in above equation, we get:

\text{Moles of ethanol}=\frac{21.8g}{46.07g/mol}=0.473mol

To calculate the enthalpy change of the reaction, we use the equation:

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat released = ?

n = number of moles = 0.473 moles

\Delta H_{rxn} = enthalpy change of the reaction  = -1235 kJ/mol = -1235\times 10^3J/mol     (Conversion factor:  1 kJ = 1000 J)

Putting values in above equation, we get:

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To calculate the heat capacity of calorimeter, we use the equation:

q=c\Delta T

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q = heat absorbed by the calorimeter = 584.16\times 10^3J

c = heat capacity = ?

\Delta T = change in temperature = T_2-T_1=62.3^oC-25^oC=37.3^oC

Putting values in above equation, we get:

584.16\times 10^3J=c\times 37.3^oC\\\\c=\frac{584.16\times 10^3J}{37.3^oC}=15.66J/^oC

Hence, the heat capacity of calorimeter is 15.66J/^oC

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