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Karolina [17]
3 years ago
5

HELP. NO FAKE ANSWERS. I WILL REPORT. I AM CONFUSED AND NEED HELP. FILL IN THE NOT FILLED BOXES POR FAVOR.

Chemistry
1 answer:
SCORPION-xisa [38]3 years ago
7 0
Question #1
Potasium hydroxide (known)
 volume used is 25 ml 
Molarity (concentration) = 0.150 M
Moles of KOH used 
           0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4) 
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l) 
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
      0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid)  concentration
    0.001875 moles × 1000/15  
        = 0.125 M

Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
      0.25 × 30/1000
        =  0.0075 moles 
Sodium Hydroxide (base)
Volume used 20 ml 
Concentration (unknown)
The equation for the reaction is 
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr   is 1 : 1
Therefore, moles of NaOH used;
                 = 0.0075 moles
NaOH concentration will be 
       = 0.0075 moles × 1000/20
       = 0.375 M

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Describe the experimental basis for believing that the nucleus occupies a very small fraction of the volume of the atom.
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Explanation:

Rutherford conducted an experiment in which he took a thin gold particle film on which he passes alpha- particles. He noticed that:

  • Most of the alpha particles get through the film and can be detected by the detector.
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He concluded that:

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A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO
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Answer: The empirical formula for the given compound is CH_2

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

C_xH_y+O_2\rightarrow CO_2+H_2O

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of CO_2=22.0g

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, \frac{12}{44}\times 22.0=6g of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.5 moles.

For Carbon = \frac{0.5}{0.5}=1

For Hydrogen  = \frac{1.0}{0.5}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is C_{1}H_{2}=CH_2

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