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tigry1 [53]
3 years ago
8

The reaction below demonstrates which characteristic of a base? Upper C U upper S upper O subscript 4 (a q) plus 2 upper N a upp

er O upper H (a q) right arrow upper C u (upper O upper H) subscript 2 (s) plus upper N a subscript 2 upper S upper O subscript 4 (a q). the ability of bases to release sodium ions into solution the ability of bases to release hydrogen ions into solution the ability of a base to react with carbonate or bicarbonate the ability of a base to react with a soluble metal salt
Chemistry
2 answers:
Anton [14]3 years ago
8 0

Answer:

the ability of a base to react with a soluble metal salt

Explanation:

just finished taking the test got 100% correct!

Good luck your awesome! :D

Gnesinka [82]3 years ago
6 0

"The ability of a base to react with a soluble metal salt" demonstrates characteristic of a base.

<u>Answer:</u> Option D

<u>Explanation:</u>

As the sodium hydroxide react with only with some metals. It is a very strong basic compound with a intense solubility in water and hygroscopic in nature. The ability of base is reflected from its potential to react with metals. In the given reaction:

\mathrm{CuSO}_{4}(a q)+2 \mathrm{NaOH}(a q) \rightarrow \mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)

\mathrm{Cu} \mathrm{SO}_{4} is an acidic property showing salt which is added with sodium hydroxide and this reaction result into a pale blue precipitate of basic copper hydroxide & sodium sulphate which is neutral salt. The reaction is a double displacement and precipitation reaction.

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What is k in the rate law equation?
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An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 6.50 g of this compound produced 9.53 g o
OverLord2011 [107]

Answer:

There were 0.216 moles of oxygen in the sample

Explanation:

Step 1: Data given

Mass of compound = 6.50 grams

Mass of CO2 = 9.53 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 3.90 grams

Molar mass H2O = 18.02 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 9.53 grams / 44.01 g/mol

Moles CO2 = 0.217 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.217 moles CO2 we have 0.217 moles C

Step 4: Calculate mass C

Mass C = 0.217 moles C *12.01 g/mol

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Step 5: calculate moles H2O

Moles H2O = 3.90 grams / 18.02 g/mol

Moles H2O = 0.216 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.216 moles H2O we have 0.433 moles H

Step 7: Calculate mass H

Mass H = 0.433 moles * 1.01 g/mol

Mass H = 0.437 grams

Step 8: Calculate mass O

Mass O = 6.50 grams - 2.61 grams - 0.437 grams

Mass O = 3.453 grams

Step 9: Calculate moles O

Moles O = 3.453 grams / 16.0 g/mol

Moles O = 0.216 moles

There were 0.216 moles of oxygen in the sample

4 0
3 years ago
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