Answer:
Root mean squared velocity is different.
Explanation:
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In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:
Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.
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Answer: C is the correct answer
Explanation:
The volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
<h3>Dilution formula</h3>
M₁V₁ = M₂V₂
Where
- M₁ is the molarity of stock solution
- V₁ is the volume of stock solution
- M₂ is the molarity of diluted solution
- V₂ is the volume of diluted solution
<h3>Data obtained from the question </h3>
- Molarity of stock solution (M₁) = 24.7 mol/L
- Volume of diluted solution (V₂) = 8 L
- Molarity of diluted solution (M₂) = 10 mol/L
- Volume of stock solution needed (V₁) = ?
<h3>How to determine the volume needed </h3>
The volume of the methanol necessary to prepare the solution can be obtained as illustrated below:
M₁V₁ = M₂V₂
24.7 × V₁ = 10 × 8
24.7 × V₁= 80
Divide both side by 24.7
V₁ = 80 / 24.7
V₁ = 3.2 L
Thus, the volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
Learn more about dilution:
brainly.com/question/15022582
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Well, considering atoms make up everything, they clearly play a very essential role.
However, this is where we see things like balanced chemical reactions and equations. It tells us how many we need for a reaction to occur, for example 4 atoms of nitrogen, etc.
Multiply first the mass given in kilograms by 1000 in order to convert it to grams.
(0.23 kg) (1000 g / 1kg) = 230 g
Then, divide the mass in gram by the molar mass of SO2
molar mass of SO2 = 64.044 g/mol
Solving for the number of moles,
(230 g)(1 mol/64.044g) = 3.59 moles
Thus, there are approximately 3.59 moles of SO2 present in 0.23 kg of SO2.