Since the equation for density is d=m/v, you can multiply both sides by m in order to get m by itself {d(v)=m}. Plug the volume and density in (1.1x7.9) and you will get 8.69. If you are using proper sig figs, then it will be 8.7g.
Answer: 2.22 litres
Explanation:
Given that,
Original pressure of gas (P1) = 200.0 kPa
[Since final pressure is given in atmosphere, Convert 200.0 kPa to atmosphere
If 101.325 kPa = 1 atm
200.0 kPa = 200.0/101.325
= 1.9738 atm]
Original volume of oxygen O2 (V1) = 25.5L
New pressure of gas (P2) = 11.2 atm
New volume of gas (V2) = ?
Since pressure and volume are given while temperature is held constant, apply the formula for Boyle's law
P1V1 = P2V2
0.9738 atm x 25.5L = 11.2 atm x V2
24.8319 atm•L = 11.2 atm•V2
Divide both sides by 11.2 atm to get V2
24.8319 atm•L/11.2 atm = 11.2 atm•V2/11.2 atm
2.217 L = V2
[Round 2.217L to the nearest hundredth as 2.22L]
Thus, the new volume of the gas at 11.2 atm of pressure is 2.22 litres
Answer:
Born-Haber Process - "Born-Haber process applies Hess's law to calculate the lattice enthalpy(<em>this is a thermodynamic term which equals to the total heat content of a system</em>), by comparing the standard enthalpy change of formation of ionic compound (from the elements) to the enthalpy required to make gaseous ions from the elements".
Explanation:
Now the calculations done for SRI2 are as follows:
- Enthalpy of Sublimation of Sr(s)= 164 kJ per mol,
- 1st ionization energy of Sr(g)=549 kJ per mol,
- 2nd ionization energy of Sr(g)=1064 kJ per mol,
- Enthalpy dissociation energy of I2(s)= 62.4 kJ per mol,
- Bond dissociation energy of I2(g)= 152.55 kJ per mol,
- 1st electron affinity of I(g)= -295.15 kJ per mol,
- Lattice energy of SrI2(s)=-1959.75 kJ per mol.
Note - the terms mentioned in the brackets are as, S- solid form, g- Gaseous form.