Answer:
Mass of sample is 99.9 g, density of ample is 11.35 
and temperature of sample is 
Explanation:
Mass is an additive property. Therefore mass of combined sample is summation of masses of two pellets.
Mass of combined sample = (37.2+62.7) g = 99.9 g
Density is an intensive property. Therefore density of combined sample of lead will be same as with density of Pb.
Density of combined sample = 11.35
Temperature is an intensive property. Therefore temperature of combined sample of lead will be same as with individual pellets.
temperature of combined sample =
When the body senses foreign substances (called antigens), the immune system works to recognize the antigens and get rid of them. B lymphocytes are triggered to make antibodies (also called immunoglobulins). These proteins lock onto specific antigens.
Answer:
Ability to conduct electricity
Melting point
Answer:
7.28 moles Ag°
Explanation:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Given 7.28 moles 7.28 moles
To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...
Limiting reactant determination:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Cu: 7.28 / 1 = 7.28
AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃
Solving Problem depends on AgNO₃; Cu will be in excess.
Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.
