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Georgia [21]
4 years ago
11

Suppose you have 15.0 ml of zinc. What volume of aluminum would have the same mass as the zinc?

Chemistry
1 answer:
Black_prince [1.1K]4 years ago
5 0

Answer:

\large \boxed{\text{41.2 cm}^{3}}

Explanation:

1. Mass of zinc

The density of zinc is 7.14 g/cm³

D = m/V

\begin{array}{rcl}7.14 & = & \dfrac{m}{15.0}\\\\m & = & 7.14 \times 15.0\\& = & \text{107.1 g}\\\end{array}\\

2. Volume of aluminium

The density of aluminium is 2.60/cm³.

\begin{array}{rcl}2.60 & = & \dfrac{107.1}{V}\\\\2.60V & = & 107.1\\V& = &\dfrac{107.1}{2.60}\\\\& = &\textbf{41.2cm}^{\mathbf{3}}\\\end{array}\\\large \boxed{\textbf{41.2 cm}^{\mathbf{3}}}\text{ of aluminium have the same mass as 15.0 cm$^{3}$ of zinc.}

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A solution is prepared by dissolving 0.23 mol of lactic acid and 0.27 mol of sodium lactate in water sufficient to yield 1.00 L
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Explanation:

The solution of the lactic acd and sodium lactate is referred to as a buffer solution.

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When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:

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3 years ago
At a particular temperature, K 3.39 for the reaction SO2(9) + NO2(9) SOs(9) +NO(9) If all four gases had initial concentrations
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Answer:

The equilibrium concentrations are:

[SO2]=[NO2] =  0.563 M

[SO3]=[NO] =  1.04 M

Explanation:

<u>Given:</u>

Equilibrium constant K = 3.39

[SO2] = [NO2] = [SO3] = [NO] = 0.800 M

<u>To determine:</u>

The equilibrium concentrations of the above gases

Calculation:

Set-up an ICE table for the given reaction

         SO2(g) + NO2(g)\rightleftharpoons  SO3(g) + NO(g)

I                0.800      0.800                                  0.800       0.800

C                -x               -x                                         +x             +x

E              (0.800-x)   (0.800-x)                             (0.800+x)   (0.800+x)

The equilibrium constant is given as:

Keq = \frac{[SO3][NO]}{[SO2][NO2]}=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}

3.39=\frac{(0.800+x)^{2}}{(0.800-x)^{2}}

x = 0.2368 M

[SO2]=[NO2] = 0.800 -x = 0.800 - 0.2368 = 0.5632 M

[SO3]=[NO] = 0.800 +x = 0.800 + 0.2368 = 1.037 M

4 0
3 years ago
An unknown amount of Al203 decomposed producing 215 g of solid aluminum. 2Al2O3=4Al+3O2 How many grams of oxygen gas should be p
natulia [17]

Answer:

191.11 grams of oxygen gas should be produced.

Explanation:

The balanced reaction is:

2 Al₂O₃ → 4 Al + 3 O₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al₂O₃: 2 moles
  • Al: 4 moles
  • O₂: 3 moles

Being the molar mass of each compound:

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  • Al: 27 g/mole
  • O₂: 32 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al₂O₃: 2 moles* 102 g/mole= 204 grams
  • Al: 4 moles* 27 g/mole= 108 grams
  • O₂: 3 moles* 32 g/mole= 96 grams

Then you can apply the following rule of three: if by stoichiometry 108 grams of aluminum are produced along with 96 grams of oxygen, 215 grams of aluminum are produced along with how much mass of oxygen?

mass of oxygen=\frac{215 grams of aluminum*96 grams of oxygen}{108grams of aluminum}

mass of oxygen= 191.11 grams

<u><em>191.11 grams of oxygen gas should be produced.</em></u>

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