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MA_775_DIABLO [31]
2 years ago
13

Preparation the buffer solution: initial pH of buffer solution: ____ Titration of a weak acid with a strong base: initial pH of

weak acid: ____ final pH of weak acid: ____ Amount of NaOH added: ____ Titration Curve for Weak Acid with a Strong Base (Paste curve here.)
Chemistry
1 answer:
Nikolay [14]2 years ago
4 0

Answer:

pH of buffer solution is 7.0

Initial pH of Weak acid is 3.27

Final pH of weak acid is 3.07

Amount of NaOH added is 1ml

Explanation:

Titration is a process in which acid and base are introduced together until a neutral solution is achieved whose pH value is near to buffer solution which is 7.0, the pH value for acid is below 7 while pH value for base is above 7.

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Express the following answers with the correct number of significant figures.<br> 3.582 ÷ 0.0131
kkurt [141]
Answer is 273

explanation:

The rule of division for significant figures is to put the answer to the least amount of significant figures presented in the problem.
0.0131 only has 3 sig figs which is 131
3.582 has 4 sig figs
The least amount is 3 in the problem
When you originally divide 3.582 and 0.0131 you get 273.4531 in your calculator

Now put it to 3 sig figs
4 is not greater than 5 so you don't need to round
273 is 3 sig figs.
So answer is 273
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3 years ago
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4. Base your answer to the following question on the elements in Group 2 on the Periodic Table. State, in terms of the number of
Zolol [24]

Answer:

The atomic radius of strontium is larger than magnesium because there are more number of shielding electrones in strontium than magnesium, which causes electron-electron repulsion, that makes the electrons in valance shell to expand. Hence it would decrease the effective nuclear charge.

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2 years ago
Please help me this is my fourth attempt.
Vaselesa [24]

Explanation:

CH4 + 4S ---> CS2 + 2H2S

4) 0.75 mol S × (1 mol CS2/4 mol S) = 0.19 mol CS2

5) 3 mol H2S × (1 mol CH4/2 mol H2S) = 1.5 mol CH4

Fe2O3 + 2Al ---> 2Fe + Al2O3

6) 25 g FeO3 × (1 mol Fe2O3/159.69 g Fe2O3) = 0.16 mol Fe2O3

0.16 mol Fe2O3 × (2 mol Al/1 mol Fe2O3) = 0.32 mol Al

0.32 mol Al × (26.98 g Al/1 mol Al) = 8.6 g Al

7) Given:

45 g Al × (1 mol Al/26.98 g Al) = 1.6 mol Al

85 g Fe2O3 ×(1 mol Fe2O3/159.69 g Fe2O3)

= 0.53 mol Fe2O3

Let's look at how much Fe each reactant will produce:

1.6 mol Al × (2 mol Fe/2 mol Al) = 1.6 mol Fe

0.53 mol Fe2O3 × (2 mol Fe/1 mol Fe2O3) = 1.1 mol Fe

Note that the given amount of Fe2O3 will give us fewer Fe. Therefore, Fe2O3 is the limiting reactant.

8) Al will produce 1.6 mol Fe × (55.845 g Fe/1 mol Fe)

= 89 g Fe

Fe2O3 will produce 1.1 mol Fe × (55.845 Fr/1 mol Fe)

= 61 g Fe

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%yield = (25 g Fe/61 g Fe) × 100% = 41%

10) If we only got 25 g Fe, then the amount of Al actually used in the reaction is

25 g Fe × (1 mol Fe/55.845 g Fe) = 0.45 mol Fe

0.45 mol Fe × (2 mol Al/2 mol Fe) = 0.45 mol Al

0.45 mol Al × (26.98 g Al/1 mol Al) = 12 g

Therefore, the leftover amount of Al is

25 g Al - 12 g Al = 13 g Al

8 0
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adell [148]

Answer:

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Explanation:

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