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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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The periodic table contains groups and periods that include the elements. For group 1 metal lithium is least likely to lose an electron.

<h3>What are group 1 metals?</h3>

Group 1 metals are the alkali metals that include, Li, Na, K, Rb, and Cs. They show the property exhibited by the metals. The chemical trends of group 1 show that cesium loses an electron more easily than the other elements.

When going down the group the tendency to lose electrons increases as the atomic radius increases. The electron gets far away from the nucleus making it easy to get removed.

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Explanation:

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