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vaieri [72.5K]
3 years ago
6

How many liters of solution will contain 1.5 moles of CaCl2 if the solution is 6.0 M CaCl2?

Chemistry
1 answer:
sashaice [31]3 years ago
6 0

Answer:

0.25

Explanation:

volume= number of moles over concentration

hence v=1.5/6

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19. When an atom of iron loses all of its 4s electrons and one of its 3d electrons, what is its charge?
Snowcat [4.5K]
Its charge would be the amount of electrons that are lost in total, which the information is not stated

7 0
3 years ago
Question 2 of 50
wolverine [178]

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

  1. Write the balanced chemical equation for the given reaction.
  2. Convert the mass of calcium carbonate into moles.
  3. Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
  4. Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

mol \ CaO \ = 0.2498 \ mol

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

\boxed {mass \ CaO \ = 14 \ g}

Learn More

  1. Learn more about stoichiometry brainly.com/question/12979299
  2. Learn more about mole conversion brainly.com/question/12972204
  3. Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

5 0
3 years ago
Calculate e°cell for a silver-aluminum cell in which the cell reaction is al(s) + 3ag+(aq) → al3+(aq) + 3ag(s) –2.46 v 0.86 v –0
tekilochka [14]
When E° cell is an electrochemical cell which comprises of two half cells.
 
So,

when we have the balanced equation of this half cell :

Al3+(aq) + 3e- → Al(s)   and E°1 = -1.66 V 

and we have  also this balanced equation of this half cell :

Ag+(aq)  + e- → Ag(s)  and E°2 = 0.8 V 

so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)

when E° = E°2 - E°1

∴E° =0.8 - (-1.66)

      = 2.46 V

∴ the correct answer is 2.46 V




6 0
3 years ago
Which of the following represents the greater concentration?<br> 1. 5 ppm<br> 2. 5%
Elanso [62]

Answer:

1

Explanation:

3 0
3 years ago
What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 - 10-147 OA) 6.767 O B) 0 465 O c) 7.000 OD) 7 233
Sergeeva-Olga [200]

Answer:

PH= 6.767     (answer is the A option)

Explanation:

first we need to correct the value in Kw at this temperature is 2.92*10^-14

so, in this case we have that:

Kw=2.92*10^-14 M²

[ H3O^+] [ H3O^+]

[H_{3}O^{+}  ] [OH^{-}  ] = Kw = 2.92*10^{-14} M^{2}   \\\\

at 40ºC

[H_{3}O^{+}  ] = [OH^{-}  ]

[H_{3}O^{+}  ]^{2} = 2.92*10^{-14} M^{2}

[H_{3}O^{+}  ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M

PH= -log10[H_{3}O^{+}  ] = -log10(1.71*10^{-7} ) = 6.767

7 0
2 years ago
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