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miskamm [114]
2 years ago
9

PROBLEMS

Chemistry
1 answer:
AlekseyPX2 years ago
7 0

Answer:

1.) 0.1 M

2.) 0.2 M

3.) 1 M

4.) Solution #3 is the most concentrated because it has the highest molarity. This solution has the largest solute to solvent ratio. The more solvent there is, the lower the concentration and molarity.

Explanation:

To find the molarity, you need to (1) convert grams NaOH to moles (via molar mass from periodic table) and then (2) calculate the molarity (via the molarity equation). All of the answers should have 1 sig fig to match the given values.

Molar Mass (NaOH): 22.99 g/mol + 16.00 g/mol + 1.008 g/mol

Molar Mass (NaOH): 39.998 g/mol

4 grams NaOH           1 mole
----------------------  x  ------------------  = 0.1 moles NaOH
                                 39.998 g

1.)

Molarity = moles / volume (L)

Molarity = (0.1 moles) / (1 L)

Molarity = 0.1 M

2.)

Molarity = moles / volume (L)

Molarity = (0.1 moles) / (0.5 L)

Molarity = 0.2 M

3.)

Molarity = moles / volume (L)

Molarity = (0.1 moles) / (0.1 L)

Molarity = 1 M

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The halpy of vaporization of H2O at 1 atm and 100 C is 2259 kJ/kg. The heat capacity of liquid water is 4.19 kJ/kg.C, and the he
Naddik [55]

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

        C_{p}_{liquid} = 4.19 kJ/kg ^{o}C

        C_{p}_{vaporization} = 1.9 kJ/kg ^{o}C

Heat of vaporization (\DeltaH^{o}_{vap}) at 1 atm and 100^{o}C is 2259 kJ/kg

        H^{o}_{liquid} = 0

Therefore, calculate the enthalpy of water vapor at 1 atm and 100^{o}C as follows.

            H^{o}_{vap} = H^{o}_{liquid} + \DeltaH^{o}_{vap}        

                                   = 0 + 2259 kJ/kg

                                   = 2259 kJ/kg

As the desired temperature is given 179.9^{o}C and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and 179.9^{o}C is calculated as follows.

             H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})

                             = 0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)

                              = 334.781 kJ/kg

Hence, enthalpy of water vapor at 10 bar and 179.9^{o}C is calculated as follows.

               H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})

             H^{D}_{vap} = 2259 kJ/kg + 1.9 \times (179.9 - 100)            

                              = 2410.81 kJ/kg

Therefore, calculate the latent heat of vaporization at 10 bar and 179.9^{o}C as follows.

       \Delta H^{D}_{vap} = H^{D}_{vap} - H^{D}_{liq}              

                         = 2410.81 kJ/kg - 334.781 kJ/kg

                         = 2076.029 kJ/kg

or,                      = 2076 kJ/kg

Thus, we can conclude that at 10 bar and 179.9^{o}C latent heat of vaporization is 2076 kJ/kg.

3 0
4 years ago
Help please. Will give brailiest
andrew11 [14]

Answer:

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Explanation:

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3 years ago
Bill is pushing a box with 10 N of force to the left, while Alice is pushing the box with 30 N of force to the right. What is th
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Answer:

Fnet = 20 Newton

Explanation:

Let the force applied by Brian be Fb.

Let the force applied by Alice be Fa.

<u>Given the following data;</u>

Force, Fb = 10N

Force, Fa = 30N

To find the net force, Fnet;

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

  • Fnet is the net force
  • Fapp is the applied force
  • Fg is the force due to gravitation

Since the two force applied to the box are acting in opposite direction, we would subtract their values to find the net force.

Fnet = Fa - Fb

Fnet = 30 - 10

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What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of acetic acid, CH3COOH
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Answer:

The molarity is 1.26

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Answer:

1.17 grams

Explanation:

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C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)

We can establish the following relations:

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The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

-59.0kJ.\frac{1molC_{2}H_{4}}{-1411kJ} .\frac{28.05gC_{2}H_{4}}{1molC_{2}H_{4}} =1.17gC_{2}H_{4}

8 0
3 years ago
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