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olchik [2.2K]
3 years ago
12

What is the M of NaOH if it takes 40 ml of NaOH to reach the equivalence point in a titration with 50 ml of 0.2 M HCl

Chemistry
1 answer:
nikdorinn [45]3 years ago
4 0

Answer:

NaOH+HCl ----->NaCl+H2O

Explanation:

number mole NaOH = number mole NaOH

M -molarity, V - volume

M(NaOH)*V(NaOH)  =  M(HCl)*V(HCl)

M(NaOH)*40 ml  =  0.2 M*50 ml

M(NaOH)=0.2 M*50 ml/40 ml=1/4 M=0.25 M NaOH

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An ore is to be analyzed for its iron content by an oxidation-reduction titration with permanganate ion. A 4.230 g sample of the
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Answer:

See explanation

Explanation:

The balanced redox reaction equation is;

8H+ + MnO4^- + 5Fe2+ ---------> Mn2+ + 5Fe3+ + 4H2O

Amount of KMnO4 reacted = 31.60/1000 * 0.05120 = 1.62 * 10^-3 moles

From the reaction equation;

1 mole of MnO4^- reacted with 5 moles of Fe2+

1.62 * 10^-3 moles will react with 1.62 * 10^-3 moles * 5/1 = 8.1 * 10^-3 moles

Mass of Fe2+ reacted = 8.1 * 10^-3 moles  *  56 g/mol

Mass of Fe2+ reacted = 0.45 g

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2 years ago
A _________________ is developed through the scientific method, and it can be modified or improved upon. It may be represented b
hodyreva [135]
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3 years ago
Sulfuric acid dissolves aluminum metal according to the following reaction:
Fiesta28 [93]

Answer:

m_{H_2SO_4}=81.7gH_2SO_4

m_{H_2}=1.67gH_2

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

2Al (s) + 3H_2SO_4 (aq)\rightarrow  Al _2(SO4)_3 (aq) + 3H_2 (g)

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2

Best regards.

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3 years ago
If u answer this correctly I’ll mark you brainliest
olga2289 [7]

Answer:

6 is the right answer I know cause I like science

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3 years ago
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