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3 years ago

Is sodium hypochlorite an oxidizing or reducing agent?

1 answer:

5 0

Sodium hypochlorite: NaClO, Bleach solution, is an oxidizing agent. Oxidizing agents include halogens, potassium nitrate, and nitric acid. Hope this helps! :)

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وزن الملي مكافئ لحامض الخليك

Natalka [10]

Answer:

hshssytdtctdyeheb

Explanation:

yye6d66d6d6dududyydydydyehwj2

3 0

3 years ago

The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) C

Kipish [7]

Answer: 72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the number of moles, we use the equation:

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

Putting in the values we get:

$\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles$

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

According to stoichiometry:

1 mole of $C_3H_8$ requires 5 moles of oxygen

0.45 moles of $C_3H_8$ require= $\frac{5}{1}\times 0.45=2.25$ moles of oxygen

Mass of $O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g$

72 grams of $O_2(g)$ are needed to completely burn 19.7 g $C_3H_8(g)$

7 0

3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

Two closed vessels contain chlorine gas at the same conditions of temperature and

wariber [46]

Answer:12 mol

Explanation: both vessels are at the same temp and pressure (and the pressure is low and/or the temperature high).

6.7mol per 1.3L = 6.7/1.3 mol/L

so in 2.33L = 6.7*2.33/1.3 = 12 mol

8 0

3 years ago

How many grams of zinc would be required to produce 9.65g of zinc hydroxide

Effectus [21]

The question is incomplete, here is the complete question:

How many grams of zinc would be required to produce 9.65g of zinc hydroxide

Zn+2MnO₂+H₂O→Zn(OH)₂+Mn₂O₃

Answer: The mass of zinc required is 6.35 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of zinc hydroxide = 9.65 g

Molar mass of zinc hydroxide = 99.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of zinc hydroxide}=\frac{9.65g}{99.4g/mol}=0.0971mol$

The given chemical equation follows:

$Zn+2MnO_2+H_2O\rightarrow Zn(OH)_2+Mn_2O_3$

By Stoichiometry of the reaction:

1 mole of zinc hydroxide is produced from 1 mole of zinc

So, 0.0971 moles of zinc hydroxide will be produced from = $\frac{1}{1}\times 0.0971=0.0971mol$ of zinc

Now, calculating the mass of zinc from equation 1, we get:

Molar mass of zinc = 65.4 g/mol

Moles of zinc = 0.0971 moles

Putting values in equation 1, we get:

$0.0971mol=\frac{\text{Mass of zinc}}{65.4g/mol}\\\\\text{Mass of zinc}=(0.0971mol\times 65.4g/mol)=6.35g$

Hence, the mass of zinc required is 6.35 grams

6 0

3 years ago