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Nina [5.8K]
3 years ago
8

You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accurately to 4 significant fi

gures. You use a simple metric ruler to determine its size and calculate the results in column A. Your partner (using a precision micrometer for the first time) obtains the results in column B. (The accepted density of aluminum is 2.702 g/cm3.)
Method A results (g/cm3) Method B results (g/cm3)

2.2 2.603

2.3 2.601

2.7 2.605

2.4 2.611

Calculate the average density for each method.
Chemistry
1 answer:
zmey [24]3 years ago
5 0

Answer:

Average density for method A = 2.4 g/cm³

Average density for method B = 2.605 g/cm³

Explanation:

In order to calculate the average density for each method, we need to add the data for each method, and then divide the result by the number of measurements (in this case is 4 for both methods):

  • For Method A:

Σ = 2.2 + 2.3 + 2.7 + 2.4 = 9.6

Average = 9.6/4 = 2.4 g/cm³

  • For Method B:

Σ = 2.603 + 2.601 + 2.605 + 2.611 = 10.420

Average = 10.420/4 = 2.605 g/cm³

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answer: 16 grams of SO₂ (2 sig figs)

check the question to see if its asked for a specific unit for mass (grams or kilograms, if they asked for kiligrams then convert 16 grams to kilograms by dividing it by 1000)


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Which of the following is the balanced equation for the
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How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?
kramer

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

  • Given mass=250g

So ,

\\ \Large\sf\longmapsto 250\times 15\%

\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}

\\ \Large\sf\longmapsto 37.5g

<u>37.5g of NaCl present in 250g of solution.</u>

8 0
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What is an intensive property? *
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This is the answer

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The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

8 0
3 years ago
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