When a substance is oxidized it is called a(n): select one:

1 answer:

4 0

When a substance is oxidized, its oxidation number increases. The reduction in oxidation number is due to lose of electron(s) to another substance in the reactant side. Substance which gets oxidize must reduce others. Thus it is a reducing agent.

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Determine the volume of 3.30 mol of a gas at 25°C and 0.995 atm.

Alexxx [7]

Answer:

V = 81.14 L

Explanation:

Given data:

Volume of gas = ?

Number of moles = 3.30 mol

Temperature of gas = 25°C

Pressure of gas = 0.995 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will convert the temperature.

25+273 = 298 K

now we will put the values in formula:

V = 3.30 mol 0.0821 atm.L/ mol.K 298 K / 0.995 atm

V = 80.74 L. atm / 0.995 atm

V = 81.14 L

4 0

3 years ago

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:

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