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DerKrebs [107]
3 years ago
12

Help!!!!! please there is no question just fill in the question marks

Chemistry
2 answers:
anzhelika [568]3 years ago
6 0
2 grams :) djdjdjdjsjwjdkndncp
Vaselesa [24]3 years ago
3 0

Answer:

32 grams

Explanation:

Ignore the other answer. Two oxygen atoms have a mass of 32 grams. You can also arrive at this answer by subtracting the given values (i.e. 36 - 4 = 32).

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What is the correct balanced chemical equation for this reaction?
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CH₄ + 2O₂ ---> CO₂ + 2H₂O
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Adding which of the following substances will decrease the solubility of calcium chloride in water: CaCl2(s) Ca2+(aq) + 2 Cl–(aq
Korvikt [17]

Answer:

Option b. Sodium chloride.

Explanation:

Consider the dissociation equilibrium of \rm CaCl_2 \; (s) in water:

\rm CaCl_2 \; (s) \rightleftharpoons Ca^{2+} \; (aq) + 2\; Cl^{-}\; (aq).

By the Le Chatelier's Principle, increasing the concentration of either product will shift the equilibrium to the left. Some \rm CaCl_2 \; (s) that was initially dissolved will precipitate out of the solution. This effect is called the common-ion effect.

For this \rm CaCl_2 \; (s) solution, the products of dissociation are

  • \rm Ca^{2+} ions, and
  • \rm Cl^{-} ions.

Adding either to the solution will trigger the common-ion effect and reduce the solubility of \rm CaCl_2 \; (s).

In the two choices,

  • Sodium chloride \rm NaCl will add \rm Na^{+} and \rm Cl^{-} ions to the solution.
  • Sodium fluoride \rm NaF will add \rm Na^{+} and \rm F^{-} ions to the solution.

\rm NaCl contains \rm Cl^{-} ions. It is capable of triggering the common-ion effect. However \rm NaF contains neither \rm Ca^{2+} ions nor \rm Cl^{-} ions. It will not trigger the common-ion effect.

6 0
3 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


3 0
4 years ago
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