Answer:
6H2 + P4→ 4PH3
Explanation:
Phosphorus has 4 in it and hydrogen has 3 in it. in order to balance it, we have to put 4 in front of phosphine so that the phosphorus on the product side has an equal amount as to the one on the reactant side.
the only one left to balance is hydrogen and so in order to balance it we put a 6 on h2 because the hydrogen in the product size becomes 12 (4 * 3).
therefore the hydrogen on the reactant side becomes 12 as well (6 * 2)
Answer:
d = 0.9 g/L
Explanation:
Given data:
Number of moles = 1 mol
Volume = 24.2 L
Temperature = 298 K
Pressure = 101.3 Kpa (101.3/101 = 1 atm)
Density of sample = ?
Solution:
PV = nRT (1)
n = number of moles
number of moles = mass/molar mass
n = m/M
Now we will put the n= m/M in equation 1.
PV = m/M RT (2)
d = m/v
PM = m/v RT ( by rearranging the equation 2)
PM = dRT
d = PM/RT
The molar mass of neon is = 20.1798 g/mol
d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K
d = 20.1798 g/22.413 L
d = 0.9 g/L
Answer: S8 + O2 → S8O2: not 100% sure
Explanation:
A balanced chemical equation happens when the quantity of the particles is required on the reactants side is equivalent to the quantity is the molecules in the items side.
Answer:
![\large \boxed{109.17 \, ^{\circ}\text{C}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B109.17%20%5C%2C%20%5E%7B%5Ccirc%7D%5Ctext%7BC%7D%7D)
Explanation:
Data:
50/50 ethylene glycol (EG):water
V = 4.70 gal
ρ(EG) = 1.11 g/mL
ρ(water) = 0.988 g/mL
Calculations:
The formula for the boiling point elevation ΔTb is
![\Delta T_{b} = iK_{b}b](https://tex.z-dn.net/?f=%5CDelta%20T_%7Bb%7D%20%3D%20iK_%7Bb%7Db)
i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.
1. Moles of EG
![\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}](https://tex.z-dn.net/?f=%5Crm%20n%20%3D%200.50%20%5Ctimes%20%5Ctext%7B4.70%20gal%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B3.785%20L%7D%7D%7B%5Ctext%7B1%20%20gal%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1000%20mL%7D%7D%7B%5Ctext%7B1%20L%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1.11%20g%7D%7D%7B%5Ctext%7B1%20mL%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%7D%7D%7B%5Ctext%7B62.07%20g%7D%7D%20%3D%20%5Ctext%7B159%20mol%7D)
2. Kilograms of water
![m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}](https://tex.z-dn.net/?f=m%20%3D%200.50%20%5Ctimes%20%5Ctext%7B4.70%20gal%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B3.785%20L%7D%7D%7B%5Ctext%7B1%20%20gal%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B998%20g%7D%7D%7B%5Ctext%7B1%20L%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20kg%7D%7D%7B%5Ctext%7B1000%20g%7D%7D%20%3D%20%5Ctext%7B8.88%20kg%7D)
3. Molal concentration of EG
![b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}](https://tex.z-dn.net/?f=b%20%3D%20%20%5Cdfrac%7B%5Ctext%7B159%20mol%7D%7D%7B%5Ctext%7B8.88%20kg%7D%7D%20%3D%20%5Ctext%7B17.9%20mol%2Fkg%7D)
4. Increase in boiling point
![\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}](https://tex.z-dn.net/?f=%5Crm%20%5CDelta%20T_%7Bb%7D%20%3D%20iK_%7Bb%7Db%20%3D%201%20%5Ctimes%200.512%20%5C%2C%20%5C%2C%20%5E%7B%5Ccirc%7D%5Ctext%7BC%7D%20%5Ccdot%20kg%20%5Ccdot%20mol%5E%7B-1%7D%20%5C%2C%20%5Ctimes%2017.9%20%5Ccdot%20mol%20%5Ccdot%20kg%5E%7B-1%7D%20%3D%209.17%20%5C%2C%20%5E%7B%5Ccirc%7D%5Ctext%7BC%7D)
5. Boiling point