The formula tells you the proportion in mols
2 mols of Cl: 1 mol of Pt
Now use the atomic masses:
Cl: 35.5 g/mol
Pt: 195 g/mol
Then you have these equation:
[2*35.5 g Cl] / 195 g Pt = x / 500 kg Pt
Solve for x.
x = (71 g Cl / 195 g Pt) * 500 kg Pt
x = 182 Kg Cl = 182 kg Cl2.
Explanation:
25. Binary ionic compounds: They are known as the compounds which contains a cation formed from a metal and an anion formed from a non-metal or a polyatomic anion. For example : Barium sulfate
Ba : +2 ( metal of group II-A)
: -2 ( sulfate)
These two ions combine to form a binary ionic compound having formula 
Binary Molecular compounds: They are known as the compounds which contain two or more non-metals that are bonded covalently. In these compounds, no ions are present because the electrons are shared by the atoms. Examples are:
: phosphorous trichloride
: sulfur dioxide
26. The general chemical formula for an acid is HX where H is the hydrogen and X is the non-metal or polyatomic ions. Examples of acids are:
HF : Hydrofluoric acid
: Sulfuric acid
The purpose of a switch in a series circuit is to make it easy to open or close the electrical circuit, turning the flow of electricity on or off. A light bulb in a series circuit is used to determine whether or not electricity is flowing.
Potassium carbonate, K 2CO 3, sodium iodide, NaI, potassium bromide, KBr, methanol, CH 3OH, and ammonium chloride, NH 4Cl, are s
slava [35]
Answer:
Potassium carbonate (K₂CO₃)
Explanation:
The compounds dissociate into ions in water, as follows:
K₂CO₃ → 2 K⁺ + CO₃⁻ ⇒ 3 dissolved particles per mole
NaI → Na⁺ + I⁻ ⇒ 2 dissolved particles per mole
KBr → K⁺ + Br⁻ ⇒ 2 dissolved particles per mole
CH₃OH → CH₃O⁻ + H⁺ ⇒ 2 dissolved particles per mole
NH₄Cl → NH₄⁺ + Cl⁻ ⇒ 2 dissolved particles per mole
Therefore, the largest number of dissolved particles per mole of dissolved solute is produced by potassium carbonate (K₂CO₃).
Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄
Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.
For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:
Kp = 
where:
P(N₂O₄) and P(NO₂) are the partial pressure of each gas.
Calculating constant:
Kp = 
Kp = 0.0104
After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.
P(N₂O₄) + P(NO₂) = 200
P(N₂O₄) = 200 - P(NO₂)
Kp = 
0.0104 = ![\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B200%20-%20P%28NO_%7B2%7D%29%20%20%7D%7B%5BP%28NO_%7B2%7D%20%29%5D%5E%7B2%7D%7D)
0.0104
+
- 200 = 0
Resolving the second degree equation:
=
= 98.7
Find partial pressure of N₂O₄:
P(N₂O₄) = 200 - P(NO₂)
P(N₂O₄) = 200 - 98.7
P(N₂O₄) = 101.3
The partial pressures are
= 98.7 MPa and P(N₂O₄) = 101.3 MPa