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Marta_Voda [28]
2 years ago
5

Check my answers please? 1. Which of the following is a correctly written thermochemical equation? NH4Cl NH4+ + Cl 2C8H18 + 25O2

16CO2 + 18H2O, H = 5,471 kJ/mol Fe (s) + O2 (g) Fe2O3 (s), H = 3,926 kJ -->C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l), H = 2,220 kJ/mol 2. How does a phase change affect a thermochemical equation? It alters the moles of reactants. -->It can affect the (delta) H value. It alters the products. It affects the balance of the equation. 3. For the reaction CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l), H = 890 kJ, what will be the change in enthalpy when 3 moles of methane react in excess oxygen?
890 kJ

-->2.67 × 103 kJ

+890 kJ

+2.67 × 103 kJ

4. In which of the following thermochemical equations would the H be considered a heat of solution?

NH4Cl (s) NH4+ (aq) + Cl (aq), H = + 14.8 kJ/mol

2C8H18 (l) + 25O2 (g) 16CO2 (g) + 18H2O (l), H = 5,471 kJ/mol

-->4Fe (s) + 3O2 (g) 2Fe2O3 (s), H = 3,926 kJ

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l), H = 2,220 kJ/mol

5. For the following reaction at STP, how much heat will be required to produce 63 L of NO? (At STP, one mole of any gas occupies 22.4 L.)

N2 (g) + O2 (g) 2NO (g), H = 180.5 kJ

+63 kJ

+127 kJ

-->+254 kJ

+508 kJ
Chemistry
1 answer:
snow_tiger [21]2 years ago
4 0
1. I think you chose the right answer, the equation has the states of the reactants and products. 2. I think you chose the right answer. 3. I think you also chose the right answer. Assuming that the Hrxn is written as kJ per mol CH4 4. Heat of solution is the enthalpy change associated with dissolving a solute in a solvent. I think the first choice is the right one. 5. I think you chose the right answer.
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Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s) what mass in grams of agcl is produced when 4.22 g of agno3 react with 7.73 g of alcl3?
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2 years ago
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Marie and Calvin dissolved 10 grams of KNO3 in 100 grams of water at 25oC. Next they added 5 grams more. Calvin told Marie that
PSYCHO15rus [73]
Hello!

Calvin told Marie that they could continue to add solute until the reached 40 grams because the solution was still unsaturated.

Unsaturated solutions are those in which the solvent (in this case water) can still dissolve more solute (in this case KNO₃) at the given pressure and temperature. This can be seen visually when adding more solute doesn't result in the presence of grains of solids that settle in the bottom of the flask. That happens because the rate of dissolving is higher than the rate of crystallization. 

Have a nice day!
8 0
2 years ago
Write the balanced half-equation describing the oxidation of mercury to hgo in a basic aqueous solution. Please include the stat
Cloud [144]

Answer:

Hg^0+2OH^-\rightarrow Hg^{2+}O+H_2O+2e^-

Explanation:

Hello,

In this case, mercury (II) oxide (HgO) is obtained via the reaction:

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Nonetheless, since it is a reaction carried out in basic solution, mercury's half-reaction only, must be:

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Best regards.

8 0
2 years ago
The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Use Ideal Gas Law (PV =
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Answer:

26.3 mL

Explanation:

Step 1:

Obtaining an appropriate gas law from the ideal gas equation.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by T

PV/T = nR

At this stage, we'll assume the number of mole (n) to be constant.

Note: R is the gas constant.

PV/T = constant.

We can thus, write the above equation as:

P1V1/T1 = P2V2/T2

The above equation is called the general gas equation.

Step 2:

Data obtained from the question. This includes the following:

Initial volume (V1) = 27.5 mL

Initial temperature (T1) = 22.0°C = 22.0°C + 273 = 295K

Initial pressure (P1) = 0.974 atm.

Final temperature (T2) = 15.0°C = 15.0°C + 273 = 288K

Final pressure (P2) = 0.993 atm

Final volume (V2) =..?

Step 3:

Determination of the final volume of the gas using the general gas equation obtained. This is illustrated below:

P1V1 /T1 = P2V2/T2

0.974 x 27.5/295 = 0.993 x V2/288

Cross multiply to express in linear.

295x0.993xV2 = 0.974x27.5x288

Divide both side by 295 x 0.993

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V2 = 26.3 mL

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4 0
2 years ago
The dipole moment (μ) of HBr (a polar covalent molecule) is 0.838D (debye), and its percent ionic character is 12.4 % . Estimate
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<span>When two electrical charges, of opposite sign and equal magnitude, are separated by a distance, a dipole is established. The size of a dipole is measured by its dipole moment (</span>μμ). Dipole moment is measured in Debye units, which is equal to the distance between the charges multiplied by the charge (1 Debye equals 3.34×10−30Cm3.34×10−30Cm). The dipole moment of a molecule can be calculated by Equation 1.11.1:

μ = qr

where

<span> <span>μ⃗ μ→ is the dipole moment vector</span> <span>qiqi is the magnitude of the ithith charge, and</span> <span>r⃗ ir→i is the vector representing the position of ithith charge.</span> </span>

 

r = μ/q

<span>r = [0.838D(3.34×10−30 C⋅m/ 1D)]/ (1.6×10−19 C) *0.124
</span> r = 1.41 x10^-10 m

 

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2 years ago
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