Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
Charge can be transferred from one object to another in three way conduction, polarization, and friction
Whenever electrons are transferred between objects, neutral matter becomes charged and three ways this can occur are referred to as conduction, polarization, and friction conduction means when there is direct contact between materials that differ in their ability to give up or accept electrons when two different materials are rubbed together, there is a transfer of electrons from one material to the other material and this causes one object to become positively charged (the electron loser) and the other object to become negatively charged (the electron gainer)
Here given object is water + balloon; balloon + wall; tape then you rub a balloon against your clothes and it sticks to the wall, you are adding a surplus of electrons (negative charges) to the surface of the balloon the wall is now more positively charged than the balloon and we tested the conductivity of deionized water positively charged
Know more about charge transferred one object to another
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Answer: Gunpowder
Explanation:
Gunpowder is made by mixing potassium nitrate with two fuels (atoms that can combine with oxygen atoms and release energy), carbon (charcoal) and sulfur. The oxygen atoms leave the nitrates and move onto the carbon and sulfur atoms, releasing a buttload of energy.
Answer:
1.089%
Explanation:
From;
ν =1/2πc(k/meff)^1/2
Where;
ν = wave number
meff = reduced mass or effective mass
k = force constant
c= speed of light
Let
ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl
ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl
The between the two is obtained from;
ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2
Therefore;
ν' - ν /ν = [meff/m'eff]^1/2 - 1
Substituting values, we have;
ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1
ν' - ν /ν = -0.01089
percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;
ν' - ν /ν * 100
|(-0.01089)| × 100 = 1.089%
