Answer:
88.8 minutes
Explanation:
Graham's law of diffusion relates rate of difusion by the following formula
Rate1 / rate 2 = √( Mass of argon / Mass of Neon)
Where rate = volume divided by time
Rate 1 = 10 ml / t1
Rate 2 = 10 ml / t2
Rate 1/ rate 2 = 10 ml / t1 ÷ 10 ml/ t2 = t2/ t1
t2/t1 = √(Mass of argon / mass of Neon) = √( 39.984/20.179)
125 / t1 = 1.4026
t1 = 125 / 1.4026 = 88.8 minutes
When we wish to convert a gas to liquid we have to either
a) decrease temperature
b) increase pressure
In case of fire extinguisher the CO2 is found to be in liquid state, this is as the CO2 is pressurized at high pressure which keeps CO2 in liquid state
the ideal pressure and temperature conditions when CO2 gas can be converted to CO2 gas
Pressure = 5 - 73 atm
Temperature = -57 to 31 degree Celsius
Carbon has the highest ionization energy as its energy 1086KJ\Mol and the rest are between 500 and 800.
<span>Calculating the moles and the moles ratio of the elements gives us the ratio of atoms in each element.
Converting the percentage of element into grams
40.25% carbon = 40.25/100 = .4025 * 100 g of carbon = 40.25g of C
6.19% hydrogen = 6.19/100 = .0619 * 100g g of hydrogen = 6.19g of H
8.94% oxygen = 8.94/100 = .0819 * 100 g of oxygen = 8.19g of O
44.62% bromine = 44.62/100 = .4462 * 100 g of bromine = 44.62g of Br
Converting the grams of element into moles
(48.38 g C) (1 mol/ 12.10 g C) = 4.028 mol C
(8.12 g H) (1 mol/ 1.008 g H) = 8.056 mol H
(53.38 g O) (1 mol/ 16.00 g O) = 3.336 mol O
(44.62g of Br)(0.012515018021626 moles) = 0.55842 mol Br
Calculating the moles ratio of elements by dividing the small number of moles of an element
4.028 mol C /0.55842 = 7.2 mol C x 5 = 36 mol C
8.056 mol H / 0.55842 = 14.42 mol H = 72 mol H
3.336 mol O / 0.55842 = 5.97 mol O = 30 mol O
0.55842 mol Br / 0.55842 = 1mol Br = 5 mol Br
So the empirical formula is (C6H12O5)6Br5</span>
<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³