Answer:
Explanation:
Assuming that temperature is constant
According to Boyle's Law, at constant temperature pressure is inversly proportional to the volume and mathematically it can be expressed as:
..........1
from the first equation after putting all the value
we get,
Basically team B would win since it is exerting a force of 900N unlike team A ( you can tell by doing 4900N minus 4000N ). It is very unbalanced.
Answer:
174.8 g/m is the molar mass of the solute
Explanation:
We must apply colligative property of freezing point depression.
ΔT = Kf . m . i
ΔT = T° freezing pure solvent - T° freezing solution (0° - (-2.34°C) = 2.34°C
Kf = Fussion constant for water, 1.86 °C/m
As ascorbic acid is an organic compound, we assume that is non electrolytic, so i = 1
2.34°C = 1.86°C/m . m
2.34°C / 1.86 m/°C = 1.26 m
This value means the moles of vitamin C, in 1000 g of solvent
We weighed the solute in 250 g of solvent, so let's calculate the moles of vitamin C.
1000 g ___ 1.26 moles
In 250 g ___ (250 . 1.26)/1000 = 0.314 moles
This are the moles of 55 g of ascorbic acid, so the molar mass, will be:
grams / mol ⇒ 55 g/0.314 m = 174.8 g/m
<u>Answer:</u> The average atomic mass of the element is 6.95 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
We are given:
Mass of isotope 1 = 7.02 amu
Percentage abundance of isotope 1 = 92.6 %
Fractional abundance of isotope 1 = 0.926
Mass of isotope 2 = 6.02 amu
Percentage abundance of isotope 2 = 7.42 %
Fractional abundance of isotope 2 = 0.0742
Putting values in equation 1, we get:
![\text{Average atomic mass of element}=[(7.02\times 0.926)+(6.02\times 0.0742)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20element%7D%3D%5B%287.02%5Ctimes%200.926%29%2B%286.02%5Ctimes%200.0742%29%5D)
Hence, the average atomic mass of the element is 6.95 amu
Molar mass of TiCl₃ = (47.9 + 35.5×3) g/mol = 154.4 g/mol
No. of moles of TiCl₃ = (380 g) / (154.4 g/mol) = 2.46 mol
1 mole of TiCl₃ contains 1 mole of Ti.
No. of moles of Ti needed = (2.46 mol) × 1 = 2.46 mol
Molar mass of Ti = 47.9 g/mol
Mass of Ti needed = (2.46 mol) × (47.9 g/mol) = 118 g
