First we determine the
moles CaCl2 present:
525g / (110.9g/mole) =
4.73 moles CaCl2 present
Based on stoichiometry,
there are 2 moles of Cl for every mole of CaCl2:<span>
(2moles Cl / 1mole CaCl2) x 4.73 moles CaCl2 = 9.47 moles Cl </span>
Get the mass:<span>
<span>9.47moles Cl x 35.45g/mole = 335.64 g Cl</span></span>
The formula used for determining gas pressure, volume and temperature interaction would be PV=nRT.
<span>• What is the temperature in Kelvins?
</span>You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15=
<span>K= 27 +273.15 = 300.15
It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.
</span>
<span>• Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
</span>In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
Using PV=nRT formula, you can derive that the temperature will be directly related to pressure. If the temperature decreased, the pressure will be decreased too.
<span> If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
</span>PV=nRT
P= nRT/V
P= 6 moles* <span>0.0821 L*atm/(mol*K) * 400K/25L= 7.8816 atm</span>
<span>1,3-cylohexadiene i synthesized starting from cyclohexane in following 4 steps.
1) Free Radical Substitution Rxn: Halogenation of cyclohexane in the presence of UV yield chlorocyclohexane.
2) Elimination Rxn: Dehydrohalogenation of chlorocyclohexane yields cyclohexene.
3) Halogenation of Cyclohexene (
Electrophillic Addition Rxn) gives 1,2-dihalocyclohexane.
4) Elemination Rxn: When dibromocyclohexane is treated with KOH and heated it gives 1,3-cyclohexadiene as shown below,</span>
Answer:
Military
Explanation:
I may be wrong but military seems most likely
Answer:

Explanation:
1. Write the skeleton equation for the half-reaction
NO₃⁻ ⟶ N₂O
2. Balance all atoms other than H and O
2NO₃⁻ ⟶ N₂O
3. Balance O by adding H₂O molecules to the deficient side.
2NO₃⁻ ⟶ N₂O + 5H₂O
4. Balance H by adding H⁺ ions to the deficient side.
2NO₃⁻ + 10H⁺ ⟶ N₂O + 5H₂O
5. Balance charge by adding electrons to the deficient side.
2NO₃⁻ + 10H⁺ + 8e⁻ ⟶ N₂O + 5H₂O
The amount of charge required to reduce 2 mol of NO₃⁻ is 8 F
