2CO + O2 ==> 2CO2
50 cc CO will require 25 cc O2 and will produce 50 cc CO2. That will leave 25 cc O2 remaining as un-reacted. Total volume of gas after reaction will be 75 cc; i.e., 50 cc CO2 + 25 cc O2 un-reacted. I don't know what residual gas means so take your pick of 75 cc total or 25 cc unreacted CO.
A volumue of 50cm3 of oxygen was exploded in 50cm3 of carbon (ii)oxide .what was the volume of the residual gas at room temperature
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Help me with that question
2CO+O2=2CO2
This equation represents the reaction between carbon monoxide (CO) and oxygen (O2) to form carbon dioxide (CO2).
Two molecules of CO react with one molecule of O2 to produce two molecules of CO2. This reaction is exothermic, meaning it produces heat.
It is a common industrial process, often used in the production of steel or in the manufacture of chemicals.
Two molecules of CO react with one molecule of O2 to produce two molecules of CO2. This reaction is exothermic, meaning it produces heat.
It is a common industrial process, often used in the production of steel or in the manufacture of chemicals.
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I'm sorry, I do not understand what you mean by "nyjgh". Could you please provide more context or clarify your request?
2 divide 2
2 divided by 2 is equal to 1.
reaction between magnseium and nitrogen
The reaction between magnesium and nitrogen is represented by the following equation:
3Mg + N2 → Mg3N2
This is a synthesis reaction, where magnesium (Mg) and nitrogen (N2) combine to form magnesium nitride (Mg3N2). The reaction requires high temperatures (above 350 degrees Celsius) and is often carried out under a blanket of inert gas, such as argon, to prevent the formation of magnesium oxide due to exposure to air.
Magnesium nitride is a white solid that has a variety of industrial uses, such as in the production of ceramics, pesticides, and rocket fuels. It can also react with water to form magnesium hydroxide and ammonia gas.
3Mg + N2 → Mg3N2
This is a synthesis reaction, where magnesium (Mg) and nitrogen (N2) combine to form magnesium nitride (Mg3N2). The reaction requires high temperatures (above 350 degrees Celsius) and is often carried out under a blanket of inert gas, such as argon, to prevent the formation of magnesium oxide due to exposure to air.
Magnesium nitride is a white solid that has a variety of industrial uses, such as in the production of ceramics, pesticides, and rocket fuels. It can also react with water to form magnesium hydroxide and ammonia gas.
some 150cm of nitrogen (ii)oxide was completely burnt in oxygen. what is the volume of the product formed
The balanced equation for the complete combustion of nitrogen (II) oxide (NO) in oxygen (O2) is:
2NO + O2 → 2NO2
This means that for every 2 moles of NO that react with 1 mole of O2, 2 moles of NO2 are produced.
To determine the volume of the product formed, we need to know the initial volume of NO in liters. Assuming standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. Therefore, 150 cm^3 of NO is equal to:
150 cm^3 × 1 L/1000 cm^3 × 1 mole/22.4 L = 0.0067 moles NO
Using the stoichiometry of the balanced equation, we can calculate the moles of NO2 produced:
2 moles NO → 2 moles NO2
0.0067 moles NO → x moles NO2
x = 0.0067 moles NO2
Now, we can convert the moles of NO2 to volume at STP using the same method:
0.0067 moles NO2 × 22.4 L/1 mole = 0.15 L
Therefore, the volume of NO2 produced when 150 cm^3 of NO is completely burned in oxygen is 0.15 L.
2NO + O2 → 2NO2
This means that for every 2 moles of NO that react with 1 mole of O2, 2 moles of NO2 are produced.
To determine the volume of the product formed, we need to know the initial volume of NO in liters. Assuming standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 L. Therefore, 150 cm^3 of NO is equal to:
150 cm^3 × 1 L/1000 cm^3 × 1 mole/22.4 L = 0.0067 moles NO
Using the stoichiometry of the balanced equation, we can calculate the moles of NO2 produced:
2 moles NO → 2 moles NO2
0.0067 moles NO → x moles NO2
x = 0.0067 moles NO2
Now, we can convert the moles of NO2 to volume at STP using the same method:
0.0067 moles NO2 × 22.4 L/1 mole = 0.15 L
Therefore, the volume of NO2 produced when 150 cm^3 of NO is completely burned in oxygen is 0.15 L.
calculate the mass of sodium oxide formed when 3.45g of sodium burns in air
The balanced equation for the combustion of sodium metal in air is:
4Na + O2 → 2Na2O
This means that for every 4 moles of Na that react with 1 mole of O2, 2 moles of Na2O are formed.
To calculate the mass of Na2O formed, we first need to determine the number of moles of sodium that reacted. We are given that 3.45 g of sodium burns in air, so we can calculate the number of moles of sodium using its molar mass:
molar mass of Na = 22.99 g/mol
moles of Na = mass/molar mass = 3.45 g/22.99 g/mol = 0.15 mol
Using the stoichiometry of the balanced equation, we can calculate the moles of Na2O formed:
4 moles Na → 2 moles Na2O
0.15 moles Na → x moles Na2O
x = 0.075 moles Na2O
Finally, we can calculate the mass of Na2O formed using its molar mass:
molar mass of Na2O = 61.98 g/mol
mass of Na2O formed = moles Na2O × molar mass = 0.075 mol × 61.98 g/mol = 4.65 g
Therefore, the mass of sodium oxide formed when 3.45 g of sodium burns in air is 4.65 g.
4Na + O2 → 2Na2O
This means that for every 4 moles of Na that react with 1 mole of O2, 2 moles of Na2O are formed.
To calculate the mass of Na2O formed, we first need to determine the number of moles of sodium that reacted. We are given that 3.45 g of sodium burns in air, so we can calculate the number of moles of sodium using its molar mass:
molar mass of Na = 22.99 g/mol
moles of Na = mass/molar mass = 3.45 g/22.99 g/mol = 0.15 mol
Using the stoichiometry of the balanced equation, we can calculate the moles of Na2O formed:
4 moles Na → 2 moles Na2O
0.15 moles Na → x moles Na2O
x = 0.075 moles Na2O
Finally, we can calculate the mass of Na2O formed using its molar mass:
molar mass of Na2O = 61.98 g/mol
mass of Na2O formed = moles Na2O × molar mass = 0.075 mol × 61.98 g/mol = 4.65 g
Therefore, the mass of sodium oxide formed when 3.45 g of sodium burns in air is 4.65 g.
A pupil analysed a commercial vinegar solution by titration and founf that 24.5cm^3 of 0.0981M sodium hydroxide solution was required for titration of 1cm^3 of vinegar. calculate the molarity of ethanoic acid in vinegar
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and ethanoic acid (CH3COOH) is:
CH3COOH + NaOH → CH3COONa + H2O
According to the equation, 1 mole of NaOH reacts with 1 mole of CH3COOH. Therefore, we can use the volume and molarity of NaOH used in the titration to calculate the molarity of CH3COOH in vinegar.
First, let's calculate the number of moles of NaOH used in the titration:
moles of NaOH = volume × molarity = 24.5 cm^3 × 0.0981 mol/cm^3 = 0.002401 mol
Since 1 cm^3 of vinegar was used, this means that the number of moles of CH3COOH in vinegar is also 0.002401 mol.
Now, let's calculate the molarity of CH3COOH in vinegar by dividing the number of moles of CH3COOH by the volume of vinegar used (in liters):
molarity of CH3COOH = moles of CH3COOH / volume of vinegar
= 0.002401 mol / (1 cm^3/1000)
= 2.401 mol/L
Therefore, the molarity of ethanoic acid in vinegar is 2.401 M.
CH3COOH + NaOH → CH3COONa + H2O
According to the equation, 1 mole of NaOH reacts with 1 mole of CH3COOH. Therefore, we can use the volume and molarity of NaOH used in the titration to calculate the molarity of CH3COOH in vinegar.
First, let's calculate the number of moles of NaOH used in the titration:
moles of NaOH = volume × molarity = 24.5 cm^3 × 0.0981 mol/cm^3 = 0.002401 mol
Since 1 cm^3 of vinegar was used, this means that the number of moles of CH3COOH in vinegar is also 0.002401 mol.
Now, let's calculate the molarity of CH3COOH in vinegar by dividing the number of moles of CH3COOH by the volume of vinegar used (in liters):
molarity of CH3COOH = moles of CH3COOH / volume of vinegar
= 0.002401 mol / (1 cm^3/1000)
= 2.401 mol/L
Therefore, the molarity of ethanoic acid in vinegar is 2.401 M.
how to prepare zinc hydroxide
To prepare zinc hydroxide, you can follow these steps:
Materials:
- Zinc sulfate heptahydrate (ZnSO4.7H2O)
- Sodium hydroxide (NaOH)
- Distilled water
- Stir bar
- Magnetic stirrer
- pH meter
Procedure:
1. Weigh out 10 grams of ZnSO4.7H2O and transfer it to a 250 mL beaker.
2. Add 100 mL of distilled water to the beaker containing ZnSO4.7H2O and stir the solution using a magnetic stir bar until the zinc sulfate is completely dissolved.
3. In another beaker, dissolve 4.5 grams of NaOH in 100 mL of distilled water.
4. Add the NaOH solution slowly to the ZnSO4 solution while stirring, until the pH of the solution reaches around 10.5 - 11.5. Use a pH meter to monitor and adjust the pH as necessary.
5. After adding the NaOH solution, continue stirring the mixture for another 30-40 minutes to ensure complete reaction.
6. Allow the mixture to settle and then filter the resultant precipitate using a Büchner funnel.
7. Rinse the precipitate with distilled water several times to remove any excess NaOH.
8. Dry the zinc hydroxide precipitate in an oven at 60-70 °C for several hours until it reaches a constant weight.
9. Store the dried zinc hydroxide in a dry, airtight container.
Note: This procedure and quantities are for educational purposes only. Before handling chemicals, ensure that you are familiar with their safety precautions and handle them in a safe manner.
Materials:
- Zinc sulfate heptahydrate (ZnSO4.7H2O)
- Sodium hydroxide (NaOH)
- Distilled water
- Stir bar
- Magnetic stirrer
- pH meter
Procedure:
1. Weigh out 10 grams of ZnSO4.7H2O and transfer it to a 250 mL beaker.
2. Add 100 mL of distilled water to the beaker containing ZnSO4.7H2O and stir the solution using a magnetic stir bar until the zinc sulfate is completely dissolved.
3. In another beaker, dissolve 4.5 grams of NaOH in 100 mL of distilled water.
4. Add the NaOH solution slowly to the ZnSO4 solution while stirring, until the pH of the solution reaches around 10.5 - 11.5. Use a pH meter to monitor and adjust the pH as necessary.
5. After adding the NaOH solution, continue stirring the mixture for another 30-40 minutes to ensure complete reaction.
6. Allow the mixture to settle and then filter the resultant precipitate using a Büchner funnel.
7. Rinse the precipitate with distilled water several times to remove any excess NaOH.
8. Dry the zinc hydroxide precipitate in an oven at 60-70 °C for several hours until it reaches a constant weight.
9. Store the dried zinc hydroxide in a dry, airtight container.
Note: This procedure and quantities are for educational purposes only. Before handling chemicals, ensure that you are familiar with their safety precautions and handle them in a safe manner.
400cm3 of gas D diffuses from porous plug in 50 seconds while 600cm3 of oxygen diffuses from the same porous plant in 30 seconds. Calculate the relative molecular mass of gas. (O = 16)
The rates of diffusion of two gases are proportional to the inverse square roots of their molar masses (Graham's Law of Effusion). Therefore, we can use the following equation to determine the relative molecular mass (Mr) of the gas D:
(r1 / r2) = sqrt(Mr2 / Mr1)
where:
- r1 and r2 are the rates of diffusion of gas D and oxygen, respectively
- Mr1 is the molar mass of oxygen (16 g/mol)
- Mr2 is the molar mass of gas D (what we are looking for)
First, let's calculate the rates of diffusion of gas D and oxygen:
rate of diffusion of gas D = volume / time = 400 cm^3 / 50 s = 8 cm^3/s
rate of diffusion of oxygen = volume / time = 600 cm^3 / 30 s = 20 cm^3/s
Now we can substitute these values into the equation and solve for Mr2:
(8 / 20) = sqrt(Mr2 / 16)
0.4 = sqrt(Mr2 / 16)
0.4^2 = Mr2 / 16
0.16 * 16 = Mr2
Mr2 = 2.56 g/mol
Therefore, the relative molecular mass of gas D is 2.56 g/mol.
(r1 / r2) = sqrt(Mr2 / Mr1)
where:
- r1 and r2 are the rates of diffusion of gas D and oxygen, respectively
- Mr1 is the molar mass of oxygen (16 g/mol)
- Mr2 is the molar mass of gas D (what we are looking for)
First, let's calculate the rates of diffusion of gas D and oxygen:
rate of diffusion of gas D = volume / time = 400 cm^3 / 50 s = 8 cm^3/s
rate of diffusion of oxygen = volume / time = 600 cm^3 / 30 s = 20 cm^3/s
Now we can substitute these values into the equation and solve for Mr2:
(8 / 20) = sqrt(Mr2 / 16)
0.4 = sqrt(Mr2 / 16)
0.4^2 = Mr2 / 16
0.16 * 16 = Mr2
Mr2 = 2.56 g/mol
Therefore, the relative molecular mass of gas D is 2.56 g/mol.
what volume of acidified potasium permanganate of concentration 0.02 mol dm^3 is decolorised by 200 cm^3 of hydrogen peroxide of concentration 0.02 mol dm^3. use the following ionic equation
To determine the volume of acidified potassium permanganate that is decolorized by hydrogen peroxide, we need to use the balanced chemical equation and the stoichiometry between the two reactants.
The balanced equation for the reaction between potassium permanganate (KMnO4) and hydrogen peroxide (H2O2) in acidic conditions can be represented as follows:
5H2O2 + 2KMnO4 + 6H2SO4 -> 5O2 + 2MnSO4 + 8H2O + K2SO4
From the equation, we can see that the stoichiometric ratio between KMnO4 and H2O2 is 2:5.
Given that the concentration of both hydrogen peroxide and potassium permanganate is 0.02 mol/dm^3 and the volume of hydrogen peroxide used is 200 cm^3, we can calculate the number of moles of H2O2 used:
moles of H2O2 = concentration x volume
= 0.02 mol/dm^3 x 0.2 dm^3
= 0.004 moles
Using the stoichiometric ratio, we can determine the number of moles of KMnO4 required:
moles of KMnO4 = 2/5 x moles of H2O2
= 2/5 x 0.004 moles
= 0.0016 moles
Finally, to find the volume of KMnO4 solution required, we can use the equation:
volume = moles / concentration
= 0.0016 moles / 0.02 mol/dm^3
= 0.08 dm^3
= 80 cm^3
Therefore, 80 cm^3 of acidified potassium permanganate solution of concentration 0.02 mol/dm^3 is required to decolorize 200 cm^3 of hydrogen peroxide solution with the same concentration.
The balanced equation for the reaction between potassium permanganate (KMnO4) and hydrogen peroxide (H2O2) in acidic conditions can be represented as follows:
5H2O2 + 2KMnO4 + 6H2SO4 -> 5O2 + 2MnSO4 + 8H2O + K2SO4
From the equation, we can see that the stoichiometric ratio between KMnO4 and H2O2 is 2:5.
Given that the concentration of both hydrogen peroxide and potassium permanganate is 0.02 mol/dm^3 and the volume of hydrogen peroxide used is 200 cm^3, we can calculate the number of moles of H2O2 used:
moles of H2O2 = concentration x volume
= 0.02 mol/dm^3 x 0.2 dm^3
= 0.004 moles
Using the stoichiometric ratio, we can determine the number of moles of KMnO4 required:
moles of KMnO4 = 2/5 x moles of H2O2
= 2/5 x 0.004 moles
= 0.0016 moles
Finally, to find the volume of KMnO4 solution required, we can use the equation:
volume = moles / concentration
= 0.0016 moles / 0.02 mol/dm^3
= 0.08 dm^3
= 80 cm^3
Therefore, 80 cm^3 of acidified potassium permanganate solution of concentration 0.02 mol/dm^3 is required to decolorize 200 cm^3 of hydrogen peroxide solution with the same concentration.
what is the volume in litres of 16g of oxygen
To calculate the volume of oxygen gas, given its mass of 16g, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, we need to convert the mass of oxygen to moles. The molar mass of oxygen (O2) is 32 g/mol, so:
moles of oxygen = mass / molar mass
= 16 g / 32 g/mol
= 0.5 moles
Assuming standard temperature and pressure (STP), the pressure is 1 atm, and the temperature is 273 K.
Now, we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
= (0.5 mol * 0.0821 L.atm/mol.K * 273 K) / 1 atm
= 11.2 L
Therefore, the volume of 16g of oxygen at STP is 11.2 liters.
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, we need to convert the mass of oxygen to moles. The molar mass of oxygen (O2) is 32 g/mol, so:
moles of oxygen = mass / molar mass
= 16 g / 32 g/mol
= 0.5 moles
Assuming standard temperature and pressure (STP), the pressure is 1 atm, and the temperature is 273 K.
Now, we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
= (0.5 mol * 0.0821 L.atm/mol.K * 273 K) / 1 atm
= 11.2 L
Therefore, the volume of 16g of oxygen at STP is 11.2 liters.
1 answer