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)
To convert from Kp to Kc, you need this formula---> Kp= Kc (RT)^Δn, where Δn= gas moles of product- gas moles of reactants. since you did not give a reaction formula, I can't calculate Δn. but all once you find it out. just plug it.
Kp= Kc (RT)^Δn------------------> Kc= Kp/[(RT)^Δn]
Kp= 5.23
R= 0.0821
T= 191 C= 464 K
Δn= ?
Kc= 5.23/ (0.0821 x 464)^Δn= ???
Answer:
HF - hydrogen bonding
CBr4 - Dispersion
NF3 - Dipole-dipole
Explanation:
Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom such as fluorine, chlorine nitrogen, oxygen etc. Hence the dominant intermolecular force in HF is hydrogen bonding.
CBr4 is nonpolar because the molecule is tetrahedral and the individual C-Br dipole moments cancel out leaving the molecule with a zero dipole moment hence the dominant intermolecular force are the dispersion forces.
NF3 has a resultant dipole moment hence the molecules are held together by dipole-dipole interaction.
Answer:
pH = -log₁₀ [H⁺]
Explanation:
pH is a value in chemistry used in to measure solution trying to determine each quality, purity, risks for health of some products, etc.
As you write in the question, [H⁺] = 10^(-pH)
Using logarithm law (log (m^(p) = p log(m):
log₁₀ [H⁺] = -pH
And
<h3>pH = -log₁₀ [H⁺]</h3>
Answer:
1.634 molL-1
Explanation:
The mol ration between NH3 and HCl is 1 : 1
Using Ca Va / Cb Vb = Na / Nb where a = acid and b = base
Na = 1
Nb = 1
Ca = 0.208 molL-1
Cb = ?
Va = 19.64 mL
Vb = 25.00mL
Solving for Cb
Cb = Ca Va / Vb
Cb = 0.208 * 19.64 / 25.0
Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)
Using the dilution equation;
C1V1 = C2V2
Initial Concentration, C1 = ?
Initial Volume, V1 = 25.00 mL
Final Volume, V2 = 250 mL
Final Concentration, C2 = 0.1634 molL-1
Solving for C1;
C1 = C2 * V2 / V1
C1 = 0.1634 * 250 / 25.00
C1 = 1.634 molL-1
