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Rama09 [41]
3 years ago
10

What is 0.0135 moles of in AlCl3 grams? Round to 3 sig figs

Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
6 0

Answer: 1.80g

Explanation:

Molar Mass of AlCl3 = 27 + (3x35.5)

= 27 + 106.5 = 133.5g/mol

Number of mole of AlCl3 = 0.0135mol

Mass = 0.0135 x 133.5= 1.80g

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Answer:

color

Explanation:

Scientists can use the color of minerals to tell them apart.

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3 years ago
An exothermic reaction occurs spontaneously ,why?​
Delvig [45]

Answer:

See below  

Step-by-step explanation:

An exothermic reaction tends to occur spontaneously because the products are more stable than the reactants.

Nature tries to get to the lowest energy state.

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3 years ago
State the difference between sugar in water and pebbles in water
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Sugar is made of molecules that are bonded together based on the positively and negatively charged areas.  They will slowly dissolve in water.  Pebbles are solids.  They will sit in water for a long time.  Though shale pebbles will break apart or fall apart.
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2 years ago
The 1995 Nobel Prize in Chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th
castortr0y [4]

Answer:

-162,5 kJ/mol

Explanation:

Cl(g) + 2O2(g) --> ClO(g) + O3(g)  ΔH = 122.8 kJ/mol (as we used the reaction in the opposite direction, it will turn the enthalpy from exothermic to endothermic)

2O3(g) --> 3O2(g)   ΔH = -285.3 kJ/mol

  Cl(g) + O2(g) --> ClO(g) + O3(g)  ΔH = 122.8 kJ

+ 2O3 (g)          --> 3O2(g)              ΔH = - 285.3 kJ

O3(g) + Cl(g)     --> ClO(g) + 2O2(g)  ΔH = 122.8 + (-285.3) = -162,5 kJ

4 0
3 years ago
A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 6-mL
scoray [572]

Answer:

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

  • The molar weight of CuSO₄·5H₂O is:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

  • The molarity of the first solution is:

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

  • Now we use the dilution factor in order to calculate the molarity in the second solution:

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

  • 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
  • 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
5 0
3 years ago
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