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3 years ago

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What is the change in enthalpy for the single replacement reaction between solid zinc and Cs2SO4 in solution to produce cesium a

nd ZnSO4? Zn + Cs2SO4 → 2Cs + ZnSO4 Given: `"Cs"_2"SO"_4(aq): DeltaH_f °= –1,400 "kJ"` `"ZnSO"_4(aq): DeltaH_f °= –1,063 "kJ"`

1 answer:

Digiron [165]3 years ago

4 0

The change in enthalpy refers to the amount of heat absorbed or heat evolved in a reaction which is carried out at constant pressure.

The change in enthalpy of a reaction is equal to the sum of formation of products minus sum of formation of reactants. It is denoted by $\Delta H$ .

The given reaction is:

$Zn + Cs_{2}SO_{4} \rightarrow 2Cs + ZnSO_{4}$

$\Delta H = \Delta H_{f}^{0}(products))-\Delta H_{f}^{0}(reactants))$

Substitute the given values $\Delta H_{f}^{0}$ (products) and $\Delta H_{f}^{0}$ (reactants) in above formula, we get

$\Delta H = -1,063 kJ -(-1,400 kJ)$

= $+ 337 kJ$

Thus, change in enthalpy for the single replacement reaction is $+ 337 kJ$

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Consider the following system at equilibrium where H° = 111 kJ/mol, and Kc = 6.30, at 723 K.

Rashid [163]

Answer:

1) The value of Kc:

C. remains the same.

2) The value of Qc:

A. is greater than Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium.

4) The concentration of N2 will:

B. decrease.

Explanation:

Hello,

In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:

1) The value of Kc:

C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

$ln(K)=\frac{\Delta _RG}{RT}$

2) The value of Qc:

A. is greater than Kc, since the reaction quotient is:

$Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}$

Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.

3) The reaction must:

B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.

4) The concentration of N2 will:

B. decrease, since less reactant is forming the products.

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8 0

4 years ago

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To titrate 45.00 milliliters of an unknown HCl sample, use 25.00 milliliters of a standard 0.2000 M NaOH titrant. What is the mo

gtnhenbr [62]

V ( HCl ) = 45.00 mL in liters : 45.00 / 1000 => 0.045 L

M ( HCl ) = ?

V ( NaOH ) = 25.00 / 1000 => 0.025 L

M ( NaOH) = 0.2000 M

number of moles NaOH :

n = M x V = 0.2000 x 0.025 => 0.005 moles of NaOH

Mole ratio:

HCl + NaOH = NaCl + H2O

1 mole HCl ---------- 1 mole NaOH
? mole HCl ---------- 0.005 moles NaOH

moles HCl = 0.005 x 1 / 1

= 0.005 moles of HCl :

M ( HCl ) = n / V

M ( HCl ) = 0.005 / 0.045

= 0.1111 M

hope this helps!

4 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

4 years ago

Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH

Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{12 g}{60 g/mol}$

= 0.2 mol

Molarity of acetic acid is calculated as follows.

Density = $\frac{mass}{volume}$

1 g/ml = $\frac{100 g}{volume}$

volume = 100 ml

Hence, molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.2 mol}{0.1 L}$

= 2 mol/l

As reaction equation for the given reaction is as follows.

$NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O$

So, moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

$x \times 1 M = 10 mL \times 2 M$ (as 1 L = 1000 ml)

x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.

6 0

4 years ago

How do models help scientists predict the polarity of molecules?

yuradex [85]

Answer:There are three main properties of chemical bonds that must be considered—namely, their strength, length, and polarity. The polarity of a bond is the distribution of electrical charge over the atoms joined by the bond. Specifically, it is found that, while bonds between identical atoms (as in H2) are electrically uniform in the sense that both hydrogen atoms are electrically neutral, bonds between atoms of different elements are electrically inequivalent. In hydrogen chloride, for example, the hydrogen atom is slightly positively charged whereas the chlorine atom is slightly negatively charged. The slight electrical charges on dissimilar atoms are called partial charges, and the presence of partial charges signifies the occurrence of a polar bond.

Explanation:

8 0

3 years ago

Read 2 more answers