Answer:
1. Which people are in the control group? The people who received the mint without the secret ingredient
(Group B) would be the control group.
2. What is the independent variable? Secret ingredient in the breath mint
3. What is the dependent variable? Amount of breath odor (or bad breath)
4. What should Mr. Krabs’ conclusion be? The breath mint with the secret ingredient appears to reduce the
amount of breath odor more than half the time, but it is not 100% effective.
5. Why do you think 10 people in group B reported fresher breath? This may be due to the placebo effect.
See if the carbon atoms are SP2 or Sp they the coplanirty is more and if its Sp3 hybridization it cant be in coplanar as Sp3 is having Td shape where as sp2 and sp are not :)
Answer:
1.72x10⁻⁵ g
Explanation:
To solve this problem we use the PV=nRT equation, where:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 25 °C ⇒ (25+273.16) = 298.16 K
And we <u>solve for n</u>:
- 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:
- 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g
The answer is: the mass of carbon is 420.6 grams.
m(C₈H₁₈) = 500 g; mass of octane.
M(C₈H₁₈) = 114.22 g/mol; molar mass of octane.
n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈).
n(C₈H₁₈) = 500 g ÷ 114.22 g/mol.
n(C₈H₁₈) = 4.38 mol; amount of octane.
In one molecule of octane, there are eight carbon atoms:
n(C) = 8 · n(C₈H₁₈).
n(C) = 8 · 4.38 mol.
n(C) = 35.02 mol; amount of carbon.
m(C) = 35.02 mol · 12.01 g/mol.
m(C) = 420.6 g; mass of carbone.
If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.
An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Given :
The ideal gas equation is given below.
n = PV/RT
n = 86126.25 x 0.0024 / 8.314 x 287
n = 0.622 / molar mass (n = Avogardos number)
Molar mass = 7.18 g
Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g
More about the ideal gas equation link is given below.
brainly.com/question/4147359
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