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Maksim231197 [3]
3 years ago
13

What is the protein that helps us identify the blood type

Chemistry
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

Antigens (more info below)

Explanation:

Antigens determine the blood type and may be either proteins or sugar molecule complexes (polysaccharides). Genes in the blood group antigen group provide instructions for the production of antigen proteins. Blood group antigen proteins serve a variety of functions within the red blood cell membrane.

hope I helped!

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Read 2 more answers
Co(g) effuses at a rate that is ______ times that of cl2(g) under the same conditions.
il63 [147K]

Answer is: the ratio of the effusion rate is 1.59 : 1.

1) rate of effusion of carbon monoxide gas = 1/√M(CO).  

rate of effusion of carbon monoxide gas = 1/√28.

rate of effusion of carbon monoxide gas = 0.189.  

2) rate of effusion of chlorine = 1/√M(Cl₂).  

rate of effusion of chlorine = 1/√70.9.  

rate of effusion of chlorine = 0.119.  

rate of effusion of carbon monoxide : rate of effusion of chlorine =

= 0.189 : 0.119 / ÷0.119.

rate of effusion of carbon monoxide : rate of effusion of chlorine = 1.59 : 1.

4 0
3 years ago
When 125.0 g of ethylene (C2H4) burns in 60.0 grams of oxygen to give carbon dioxide and water, how many grams of CO2 are formed
gizmo_the_mogwai [7]

Answer:

             66 g of CO₂

Solution:

The Balance Chemical Reaction is as follow,

                             C₂H₂  +  5/2 O₂    →    2 CO₂  +  H₂O

Or,

                             2 C₂H₂  +  5 O₂    →    4 CO₂  +  2 H₂O    -------  (1)

Step 1: Find out the limiting reagent as;

According to Equation 1,

            56.1 g (2 mole) C₂H₂ reacts with  =  160 g (5 moles) of O₂

So,

                  125 g of C₂H₂ will react with  =  X g of O₂

Solving for X,

                      X =  (125 g × 160 g) ÷ 56.1 g

                      X =  356.5 g of O₂

It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.

Step 2: Calculate Amount of CO₂ produced as;

According to Equation 1,

              160 g (5 mole) O₂ produces  =  176 g (4 moles) of CO₂

So,

                  60.0 g of O₂ will produce  =  X g of CO₂

Solving for X,

                      X =  (60.0 g × 176 g) ÷ 160 g

                      X =  66 g of CO₂

8 0
3 years ago
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