I haven't done these in some time, so I'm not sure if they are 100% right.
1) 6.3 moles of H2( 2 mol of NH3 / 3 mol of H2)= 4.2 mol of NH3
6.3 moles of H2( 2 mol of NH3/ 3 mol of H2)(17.04 g of NH3/1 mol NH3)= 71.57 g of NH3
2) 2.5 moles of N2(2 mol of NH3/1 mol of N2)= 5 moles of NH3
2.25 moles of N2(2 mol of NH3/ 1 mol of N2)(6.02x10^23 particles/ 1 mol of NH3)= 3.01x10^24 particles of NH3
3) 425 g of NH3(1 mol of NH3/17.04 g NH3)= 24.9 moles of NH3
425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)(28.02 g of N2/1 mol N2)= 349 g of N2
425 g of NH3(1 mol of NH3/17.04 g of NH3)(1 mol of N2/2 mol of NH3)= 12.5 mol of N2
4) 10 moles NH3(3 moles of H2/2 moles of NH3)= 15 moles H2
10 moles NH3(3 mol of H2/2 mol of NH3)(2.02 g of H2/1 mol of H2)= 30.3 g of H2
30.3 g = .0303 liters of H2
Answer:
The geometric shape of a CH4 molecule distributes the charges symmetrically.
Explanation:
The dipole moment is ruled by the compound's geometry and elements electronegativity difference.
Arranging the forces acording to the elements electronegativity difference withing the molecule geometry will yeld the dipole moment.
If the forces cancel out, it is a nonpolar compound.
If the forces does not cancel out it is a polar compound.
Picture of such analysis for methane is attached.
Answer:
See the answer below
Explanation:
The cross that would produce the maximum number of offspring with green sea pods would be:
1. GG x GG, or
2. GG x gg
The first one would produce all green offspring with homzygous genotypes as follows:
GG x GG
GG GG GG GG
The second cross would produce all green offspring with heterozygous genotypes as follows:
GG x gg
Gg Gg Gg Gg
Answer:
A potential
Explanation:
potential energy is a kind of energy which rest or stays in one position
Answer:
The concentration of the unknown acid (HA) is 0.434M
The molar mass of HA is 13.3g/mole
Explanation:
DETERMINATION OF MOLARITY OF THE UNKNOWN ACID
CaVa/CbVb = Na/Nb
From the equation of reaction and at equivalence point, Na = Nb = 1
Therefore, CaVa = CbVb
Va (volume of acid solution) = 20mL = 20/1000 = 0.2L
Cb (concentration of KOH) = 0.715M
Vb (volume of KOH) = 12.15mL
Ca (concentration of acid) = CbVb/Va
Ca = 0.715M × 12.15mL/20mL = 0.434M
DETERMINATION OF MOLAR MASS OF HA
Number of moles of acid = concentration of acid × volume of acid solution in liters = 0.434 × 0.2 = 0.0868mole
Molar mass of HA = mass/number of moles = 1.153g/0.0868mole = 13.3g/mole
