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mote1985 [20]
2 years ago
15

If 2.50 mol of copper and 5.50 mol of silver nitrate are available to react by single replacement, identify the limiting reactan

t
Chemistry
1 answer:
PIT_PIT [208]2 years ago
7 0

Answer:

The limiting reactant is the copper

Explanation:

Moles are the standard unit at which you can compare values, and copper has fewer moles than Silver nitrate, so in a reaction, it would be used up first, leaving Silver nitrate in excess.

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The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−
barxatty [35]

Answer:

The reaction will be  non spontaneous at these concentrations.

Explanation:

AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)

Expression for an equilibrium constant K_c:

K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]

Solubility product of the reaction:

K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}

Reaction between Gibb's free energy and equilibrium constant if given as:

\Delta G^o=-2.303\times R\times T\times \log K_c

\Delta G^o=-2.303\times R\times T\times \log K_{sp}

\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]

\Delta G^o=69,117.84 J/mol=69.117 kJ/mol

Gibb's free energy when concentration [Ag^+] = 1.0\times 10^{-2} M and [Br^-] = 1.0\times 10^{-3} M

Reaction quotient of an equilibrium = Q

Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}

\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)

\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])

\Delta G=40.588 kJ/mol

  • For reaction to spontaneous reaction:  \Delta G.
  • For reaction to non spontaneous reaction:  \Delta G>0.

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0
3 years ago
Rewrite each equation below with the delta h value included with either the reactants or he products , and identify the reaction
grin007 [14]
I will take a stab at it, but there are not equations, did you forget them?
5 0
3 years ago
Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to t
myrzilka [38]

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol

Moles of methane gas = n_1=\frac{8.00 g}{16 g/mol}=0.5 mol

Moles of ethane gas  =n_2=\frac{18.0 g}{30 g/mol}=0.6 mol

Moles of propane gas = n_3

n=n_1+n_2+n_3

2.22=0.5 mol +0.6 mol+ n_3

n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol

Mole fraction of methane =\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}

\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252

Similarly, mole fraction of ethane and propane :

\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703

\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

Partial pressure of methane gas:

p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm

Partial pressure of ethane gas:

p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm

Partial pressure of propane gas:

p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = n_1=\frac{37.8 g g}{28g/mol}=1.35 mol

Moles of oxygen gas  =n_2=\frac{62.2 g}{32 g/mol}=1.94 mol

Mole fraction of nitrogen=\chi_1=\frac{n_1}{n_1+n_2}

\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103

Similarly, mole fraction of oxygen

\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897

Partial pressure of each gas can be calculated by the help of Dalton's' law:

p_i=P\times \chi_1

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg

Partial pressure of oxygen gas:

p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg

3 0
3 years ago
"calculate the ratio of the velocity of hydrogen molecules to the velocity of carbon dioxide molecules at the same temperature"
Ne4ueva [31]

Answer: 1:4.69

Explanation:

The ratio can be expressed as:

Ua/Ub= √(Mb/Ma)

Where Ua/Ub is the ratio of velocity of hydrogen to carbon dioxide and Ma is the molecular mass of hydrogen gas= 2

Mb is the molecular mass of CO2 = 44

Therefore

Ua/Ub= √(44/2)

Ua/Ub = 4.69

Therefore the ratio of velocity of hydrogen gas to carbon dioxide = 1:4.69

which implies hydogen is about 4.69 times faster than carbon dioxide.

8 0
3 years ago
AB G C F G Y OBLAD
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