Answer:
There are 0,89 moles of nitrous oxide gas in the balloon.
Explanation:
We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:
PV= nRT ---> n= PV/RT
n= 1,09 atm x 20,0 L /0.082 l atm / K mol x 298 K
<em>n= 0,89212637 mol</em>
The phrase that describes the energy of collision is D. kinetic energy transferred when billiard balls hit each other.
<h3>What is collision?</h3>
collision can be regarded as coming together of one solid or direct impact to the other such as car collided with a tree.
Therefore, kinetic energy transferred when billiard balls hit each other is an example.
Learn more about collision at;
brainly.com/question/11352260
<h3>Answer:</h3>
The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.
<h3>Explanation:</h3>
Lets take start with the melting point of both compounds.
n-Butane = - 140 °C
Trimethylamine = - 117 °C
Intermolecular Forces in n-Butane:
As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.
Intermolecular Forces in Trimethylamine:
Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.
Answer:
533.33 mg quinine
Explanation:
First we<u> calculate the concentration of the diluted sample</u>, using the values obtained by the standard:
- 288 * 100 ppm / 180 = 160 ppm
Now we use the dilution factors to <u>calculate the concentration of quinine in the original sample</u>:
- 160 ppm * 100mL/15mL = 1066.67 ppm
ppm can be defined as <u>mg of quinine</u>/L solvent:
500 mL solution ⇒ 500/1000 = 0.5 L
- 1066.67 ppm * 0.5L = 533.33 mg quinine
