Depending upon the clumping reaction with anti A , anti B and anti Rh antibodies the blood types are determined.
Explanation:
Agglutination (clumping) will occur when blood that contains the particular antigen is mixed with the particular antibody.
A+ have Agglutination with Anti-A ,Anti-Rh and No agglutination with Anti-B.
A- have Agglutination with Anti-A and No agglutination with Anti-B and Anti-Rh.
B+ have Agglutination with Anti-B Anti-Rh and No agglutination with Anti-A.
B- have Agglutination with Anti-B and No agglutination with Anti-B and Anti-Rh.
Rh+ have Agglutination with Anti-A and Anti-Rh and No agglutination with Anti-B.
Rh- have No Agglutination with Anti-A and Anti-B and Anti-Rh.
<span><span>When water vapor condenses, 2260 joules/gram heat energy will be released into the atmosphere.
To add, </span>heat energy<span> <span>(or </span>thermal energy<span> or simply </span>heat) is defined as a form of energy<span> which transfers among particles in a substance (or system) by means of kinetic </span>energy<span> of those particles. In other words, under kinetic theory, the </span>heat<span> is transferred by particles bouncing into each other.</span></span></span>
The most likely answer is the boiling point and freezing point of water. The Celsius scale starts at the freezing point of water (0°C) and than scaled so that 100°C fell on the boiling point of water.
I hope this helps. Let me know if anything is unclear.
Answer:
The mass of C2H2 in the mixture is 0.56gram using the ratio of carbon in the products contributed by the C2H2.
Explanation:
The balanced equation for the reaction is: C3H8 + 2C2H2 + 10O2 >> 7CO2 + 6H2O.
From the reaction, we know that the oxygen was in excess, this will make the Carbon sources the limiting agents in the reaction. The details of the reaction showed that the ratio of water to the carbon dioxide is 1.6:1. This also means that the expected mole of carbon dioxide will be 7/1.6, which is 3.75moles.
The individual balanced equation of reaction is:
C3H3 +5O2 >> 3CO2 + 4H2O
and 2C2H2 + 5O2 >>4CO2 + 2H2O. From this one can quickly tell that the propane is in sufficient supply as it produces 3 moles of CO2 out of the expected 3.75 moles obtained above. Leaving 0.75moles of CO2 to the ethyne.
The mass of ethyne in the mixture will therefore be: 0.75/3.75 X 2.8 = 0.56g.
