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jeka94
2 years ago
6

Chemistry help!

Chemistry
1 answer:
Delvig [45]2 years ago
3 0

Answer: C, B, A

Explanation:

C is the initial solution, because naoh has not been added yet

B is the midpoint of the titration.  naoh has been added to the solution, but it has not fully reacted yet. You can tell that this one is the midpoint because there is still HF- in the diagram, which is not one of the products formed in the reaction.

A is the endpoint because the diagram shows that all products are fully formed. There is an increased amount of Na+ and H2O in the diagram and no HF- left.  

PS. I tested this answer on the concentration of acetic acid post-lab (from mcgraw hill) earlier today and it said this was the right answer :) hope this helps

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Fossil fuels is the answer
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3 years ago
How many significant figures are in the number 020.310
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5

Explanation:

Sorry, don't have one

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3 years ago
Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl
anzhelika [568]

Answer:  ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

CH_4(g)+Cl_2(g)\rightarrow CCl_4(l)+HCl(g)

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\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]

\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]

where,

n = number of moles

Now put all the given values in this expression, we get

\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]

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4 0
2 years ago
When 21.45 g of KNO3 was dissolved in water in a calorimeter, the temperature fell from 25.00°C to 14.14 °C. If the heat capacit
pashok25 [27]

25.9 kJ/mol. (3 sig. fig. as in the heat capacity.)

<h3>Explanation</h3>

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\text{KNO}_3\;(s) \to \text{KNO}_3\;(aq).

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M(\text{KNO}_3) = 39.098 + 14.007 + 3\times 15.999 = 101.102\;\text{g}\cdot\text{mol}^{-1}.

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\displaystyle n = \frac{m}{M} = \frac{21.45}{101.102} = 0.212162\;\text{mol}.

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