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Vedmedyk [2.9K]
4 years ago
8

In a labratory activity the density of a sample of vanadium is determined to be 6.9g/cm^3 at room temperature. what is the perce

nt error for the determined value?
answers:
a : 0.15%
b : 0.87%
c : 13%
d : 15%

PLEASE EXPLAIN! please
Chemistry
1 answer:
Snowcat [4.5K]4 years ago
7 0

Answer:

c : 13%

Explanation:

Data Give:

Experimental density of vanadium = 6.9 g/cm³

percent error = ?

Solution:

Formula used to calculate % error

  % error = [experimental value -accepted value/accepted value] x 100

The reported accepted density value for vanadium = 6.11 g/cm³

Put value in the above equation

                  % error = [ 6.9 - 6.11 / 6.11 ] x 100

                  % error = [ 0.79 / 6.11 ] x 100

                  % error = [ 0.129] x 100

                  % error = 12.9

Round to the 2 significant figure

% error = 13 %

So, option c is correct            

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Answer:

The correct answer is option B.

Explanation:

3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2

Moles of NBr_3 = 40 mol

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According to reaction, 3 moles of NaOH reacts with 2 moles NBr_3

Then ,48 moles of NaOH will reacts with:

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Then ,40 moles of NaBr_3 will reacts with:

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As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

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3 years ago
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vlabodo [156]

Answer:

D. C₄H₁₀ and C₂H₅

<h2>What is a empirical formula?</h2>

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Since an empirical formula indicates the ratio (proportions) of the elements in the compound, it can be used, along with the molar weight of the compound, to determine the molecular formula.

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<u>Answer:</u> The volume of acid should be less than 100 mL for a solution to have acidic pH

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To calculate the volume of acid needed to neutralize, we use the equation given by neutralization reaction:

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where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH

We are given:

n_1=1\\M_1=3.00M\\V_1=?mL\\n_2=1\\M_2=3.00M\\V_2=100mL

Putting values in above equation, we get:

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