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zmey [24]
4 years ago
14

You know the concentration of your unknown hcl solution is approximately 0.10 m and you would like to use 40. ml to reach the en

d point of the titration. what mass of tham should be used to make the 250 ml tham solution?
Chemistry
1 answer:
allochka39001 [22]4 years ago
5 0

We know the volume of acid×strength of acid = volume of base×strength of base for the neutralization of a acid base titration. Here the volume of acid is 40.0mL and strength of acid is 0.10m (m = molality).

The tham is the tromethamine. The volume of tham in the titration is 250mL.The strength of the tromethamine in the titration with hydrochloric acid (HCl) can be determined as- 40.0×0.10=s(tham)×250. The s(tham) is the strength of the solution which is 0.016m.

Now 1m tham solution is prepared by dissolving 121.14g (as the molecular weight of tham is 121.14g/mole) of tham in 1L of water. Thus to prepare 0.016m solution 121.14×0.016=1.938g of tham have to dissolve in 1000mL of water. Thus in 250mL of tham solution 1.938/4 = 0.484g of tham have to dissolve. Thus the mass of the tham present in the solution is 0.484g.  

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What is the ph of a buffer that consists of 0.45 m ch3cooh and 0.35 m ch3coona? ka = 1.8 × 10–5?
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ck(CH₃COOH) = 0.45 M.

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<span>pKa = -logKa.
</span>pKa = -log(1.8·10⁻⁵) = 4.75.
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8 0
4 years ago
WILL GIVE THE BRAIN THING-
Kruka [31]

Answer:

The products are Al(NO₃)₃(aq) and H₂(g).

Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + 1.5 H₂(g)

Explanation:

Let's consider the reaction between aluminum and nitric acid. This is a single replacement reaction, in which Al replaces H in HNO₃ to form aluminum nitrate. The unbalanced reaction is:

Al(s) + HNO₃(aq) ⇒ Al(NO₃)₃(aq) + H₂(g)

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Al(s) + 3 HNO₃(aq) ⇒ Al(NO₃)₃(aq) + 1.5 H₂(g)

5 0
3 years ago
"how many grams of h2 are needed to produce 13.09 g of nh3
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MH₂: 2 g/mol
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