Answer is: The atomic size of the chlorine ion is larger than the size of the chlorine atom.
Covalent radii of chlorine atom (Cl) is 0.099 nm and ionic radii of chlorine anion (Cl⁻) is 0.181 nm.
Difference between an chlorine atom and chlorine anion is the number of electrons that surround the nucleus.
Chlorine atom has 17 electrons and chlorine anion has 18 electrons.
The molecular formula of the compound is C12H15O3 hence the molar mass of the compound is 207 g/mol.
We need to obtain the number of moles of carbon, hydrogen and oxygen in the compound;
Carbon = 24.91 g/44g/mol × 1 mole of carbon = 0.566 moles
Mass of carbon = 0.566 moles × 12 g/mol = 6.792 g
Number of moles of hydrogen = 6.522 g/18 g/mol × 2 moles = 0.725 moles
Mass of hydrogen = 0.725 moles × 1 g/mol = 0.725 g
Mass of oxygen = 10 - (6.792 g + 0.725 g) = 2.483 g
Number of moles of oxygen = 2.483 g/16 g/mol = 0.155 moles
Now we must divide through by the lowest number of moles;
C - 0.566/0.155 H - 0.725/0.155 O - 0.155/0.155
C - 4 H - 5 O - 1
The simplest formula is C4H5O Recall that the molar mass of the compound lies between 150.0 and 220.0 g/mol
4(12) + 5(1) + 16 = 69
Hence; n = 3 and the molecular formula of the compound is C12H15O3
The molar mass of the compound is; 12(12) + 15(1) + 3(16) = 207 g/mol
Learn more: brainly.com/question/15180604
Here's the balanced equation for given Double displacement reaction ~
The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )
Answer:n = PV/RT = 0.923 atm x 0.250 L / (0.082057 x 290.75 K)
. n = 0.00967. mole 0.00967 mole x 6.22x10^23 molecules/mole = 6.02x10^21 molecules of gas
Explanation:
Answer:
We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the water is 55°C and the initial temperature is 20.0°C, ΔT is as follows:
ΔT = Tfinal − Tinitial = 55.0°C − 20.0°C = 35.0°C
given the specific heat of water as 1 cal/g·°C. Substitute the known values into heat = mcΔT and solve for amount of heat:
= heat=(75.0 g)(1 cal/ g· °C )(35.0°C) =
= 75x1x35=2625 cal
