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8_murik_8 [283]
3 years ago
6

Evidence for past events in Earth's ancient history are provided by:

Chemistry
1 answer:
g100num [7]3 years ago
7 0
Rocks and the fossils within them!
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Why are most organic compounds nonconducting and insoluble in water?
maksim [4K]
I think the correct answers from the choices listed above are the second and the last options. Organic compounds are nonducting and insoluble in water because of the bonds that these compounds contain. Most organic compounds don't dissolve in water because they are nonpolar. They don't conduct electricity because they are covalent, not ionic.
4 0
4 years ago
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How many moles of bromine atoms are in 2.60×10^2 grams of bromine?
AlladinOne [14]
Your answer is 3.25 moles of Bromine

4 0
4 years ago
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and oxalic acid (H2C2
belka [17]

Answer:

Oxalic acid is a dicarboxylic acid and forms sodium salt with NaOH and water

Explanation:

4 0
4 years ago
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write the symbols for the ions that form when potassium and iodine react to form the ionic compound potassium iodide.
Bingel [31]
I believe it isK+  + I-=KI
3 0
3 years ago
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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of sodium sulfa
notka56 [123]

Answer:

27%

Explanation:

Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

2NaOH+H_2SO_4-->Na_2SO_4+2H_2O

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4

Finally, the percent yield turns out into:

Y=\frac{1.92g}{7.1g} *100

Y=27.0%

Best regards.

6 0
3 years ago
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