How are valence electrons different from other electrons

Explain how you were able to use your knowledge of how different types of blood react with anti A, anti B, and anti Rh antibodie

dybincka [34]

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.

3 0

3 years ago

Protons and neutrons are composed of what?<br> A electrons<br> B.quarks<br> C.ions<br> D.molecules

leonid [27]

B. Quarks is the answer.

6 0

3 years ago

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10) Give two example of where you would find elements in our daily lives.

Phoenix [80]

Carbon is found in oil and gas.
Aluminum a light metal used in making pots and pans.
Bromine is used in photography.

6 0

3 years ago

What is the pH of 1.0M HI solution?

ozzi

Answer:

0

Explanation:

Since HI is a strong acid, the amoung of Hydrogen ions produced by it will be the same molar as the reactant. The negative log of the concentration will reveal that the pH is 0.

7 0

3 years ago

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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago