Lack of cell walls no chloroplasts...hoped that helped
Answer:
http://www.khalidshadid.com/uploads/3/9/2/0/3920808/raymond_chang-chemistry_10th_edition.pdf
Explanation:
http://www.khalidshadid.com/uploads/3/9/2/0/3920808/raymond_chang-chemistry_10th_edition.pdf
Answer:
1. 0.0154mole of PbS
2. Double displacement reaction
Explanation:
First, let write a balanced equation for the reaction. This is illustrated below:
Pb(CH3COO)2 + H2S —> PbS + 2 CH3COOH
Molar Mass of Pb(CH3COO)2 = 207 + 2(12 + 3 + 12 + 16 +16) = 207 + 2(59) = 207 + 118 = 325g
Mass of Pb(CH3COO)2 = 5g
Number of mole = Mass /Molar Mass
Number of mole of Pb(CH3COO)2 = 5/325 = 0.0154mole
From the equation,
1mole of Pb(CH3COO)2 produced 1mole of PbS.
Therefore, 0.0154mole of Pb(CH3COO)2 will also produce 0.0154mole of PbS
2. The name of the reaction is double displacement reaction since the ions in the two reactants interchange to form two different products
When an electron quickly occupies an strength state increased than its ground state, it is in an excited state. An electron can end up excited if it is given greater energy, such as if it absorbs a photon, or packet of light, or collides with a close by atom or particle.
Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
