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Lera25 [3.4K]
2 years ago
9

5. The heat of fusion of ice is 333.5J/g. The entropy change for the water when freezing 5.0 g of water at 0°C

Chemistry
1 answer:
lara [203]2 years ago
7 0

Answer:

(e) -6.1 J/K

Explanation:

Step 1: Given data

  • Heat of fusion of ice (ΔH°fus): 333.5 J/g
  • Mass of water (m): 5.0 g

Step 2: Calculate the heat (Qfreezing) required to freeze 5.0 g of water

We will use the following expression.

Qfreezing = -ΔH°fus × m

Qfreezing = -333.5 J/g × 5.0 g = -1.7 × 10³ J

Step 3: Calculate the entropy change (ΔS°) at 0 °C (273.15 K) and 1 atm

We will use the following expression.

ΔS° = Qfreezing/T

ΔS° = -1.7 × 10³ J/273.15 K = -6.1 J/K

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Gold is currently trading at very high price. Suppose that gold is selling for around $1860/ounce. How
Ugo [173]

Answer:

The answer is "3.81041978"

Explanation:

\to 1 \ OZ= 28.349523125 \ grams\\\\

              =28.349523125\times {1000} \ miligrams\\\\= 28349. 5231  \ miligrams\\

In 1860 = 28349.5231 \ miligrams\\

\to In \$ \ 1 = \frac{28349.5231}{1860} \ \ miligrams\\

             = 15.2416791 \ miligrams

\ In \  1 \ quarter =  \$ \ 0.25

\to \$ \ 0.25 =  15.2416791  \times 0.25 \  miligrams\\

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3 0
3 years ago
The hydrochloride form of cocaine has a solubility of 1.00 g in 0.400 mL water. Calculate the molarity of a saturated solution o
Tanya [424]

Answer:

The molarity of the solution is 7.4 mol/L

Explanation:

From the question above

0.400 ml of water contains 1.00 g of hydrochloride form of cocaine

Therefore 1000 ml of water will contain x g of hydrochloride form of cocaine

                    x = 1000 / 0.400

                    x = 2500 g

2500g of hydrochloride form of cocaine is present in 1000 ml of water.

Mole of hydrochloride form of cocaine = mass /molar mass of hydrochloride

Mole of hydrochloride form of cocaine = 2500/339.8

                                                                = 7.4 mol

Molarity = mol/ volume in liter (L)

molarity = 7.4 /1

Molarity = 7.4 mol/L

6 0
3 years ago
Read 2 more answers
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

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3 years ago
What is the mass, in grams, of a pure gold cube that has a volume of 4.20cm^3 ?
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3 years ago
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