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Bas_tet [7]
3 years ago
14

A thermometer is in a beaker of water. Which statement best explains why the thermometer reading initially increases when LiBr(s

) is dissolved in the water?
(1) The entropy of the LiBr(aq) is greater than the entropy of the water.
(2) The entropy of the LiBr(aq) is less than the entropy of the water.
(3) The dissolving of the LiBr(s) in water is an endothermic process.
(4) The dissolving of the LiBr(s) in water is an exothermic process.
Chemistry
1 answer:
fiasKO [112]3 years ago
8 0
<span>(4) The dissolving of the LiBr(s) in water is an exothermic process</span>
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How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2S
Rama09 [41]

Answer: 2.796 grams

Explanation:

Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles

\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles

According to stoichiometry:

1 mole of BaCl_2 reacts with 1 mole of Na_2SO_4

0.012 moles of BaCl_2 will react with=\frac{1}{1}\times 0.012=0.012moles of Na_2SO_4

Thus BaCl_2 is the limiting reagent as it limits the formation of product. Na_2SO_4  is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of BaCl_2 produces 1 mole of BaSO_4

0.012 moles of BaCl_2 will produce=\frac{1}{1}\times 0.012=0.012moles of BaSO_4

Mass of BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g

Thus 2.796 grams of BaSO_4  are produced.

6 0
4 years ago
What is the molarity of a 250ml KCl solution made by diluting 175mL of a 3.00 M solution?
Crank

Answer:

2.1 M

Explanation:

The dilution equation is M_{s} V_{s}  = M_{d} V_{d}.

M_{s} = the molarity of the sock solution

V_{s} = the volume of the sock solution

M_{d} = the molarity of the diluted solution

V_{d} = the volume of the diluted solution

The stock solution would be what is doing the diluting, so "175 mL of a 3.00 M solution". So M_{s} = 3.00 M. Then: Converting 175 mL to liters: 175 mL * \frac{1 L}{1000mL} = 0.175 L (This is V_{s})

And converting 250 mL KCl to liters: 250 mL * \frac{1 L}{1000mL} = 0.250 L (This will be V_{d})

Then, we plug in our given into the dilution equation, resulting in:

3.00 M * 0.175 L = M_{d} * 0.250 L (divide both sides by 0.250 L, in order to get M_{d} by itself)

\frac{3.00 M * 0.175 L}{0.250 L} = M_{d}

M_{d} = 2.1 M

So, the molarity of 250 mL KCl made by diluting 175 mL of a 3.0 M solution would be 2.1 M (mol/L).

Hopefully this helped you understand the topic a little bit better. I just finished molarity and dilutions in Chemistry last week. Good luck!

4 0
3 years ago
Pls help ill mark as brainliest :)
Alina [70]

Answer:

I think its a double reaction

Explanation:

4 0
4 years ago
Which type of bond is found in many carbon-to-carbon bonds in canola oil, but very few carbon-to-carbon bonds in butter? a.)C–C
PtichkaEL [24]
B.)C=C is the right answer.
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8 0
4 years ago
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6 0
3 years ago
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