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Natali5045456 [20]
2 years ago
11

Help me Plz, I am really confused

Chemistry
1 answer:
avanturin [10]2 years ago
5 0
O is a blood donor
AB receiver
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How many moles of glucose, C H O , can be "burned" biologically when 18.2 mol of oxygen is available? C H O (s) + 6O (g) --->
Feliz [49]
Answer:
              547.7 g of C₆H₁₂O₆

Solution:
               The balance chemical equation is as follow,

                           C₆H₁₂O₆  +  6 O₂    →    6 CO₂  +  6 H₂O

According to equation,

                         6 moles of O₂ burns  =  180.56 g of C₆H₁₂O₆
So,
                18.2 moles of O₂ will burn  =  X g of C₆H₁₂O₆

Solving for X,
                      X  =  (18.2 mol × 180.56 g) ÷ 6 mol

                      X  =  547.7 g of C₆H₁₂O₆
5 0
3 years ago
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Bodies at _____ tend to remain at rest.
Klio2033 [76]

Answer:

C.

Explanation:

Newton's 3rd law of motion: Inertia. It states that bodies at rest tend to stay at rest.

8 0
2 years ago
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A polypeptide in its native conformation has weak interactions between its R groups. However, when that same polypeptide is dena
sasho [114]

Answer:

A protein is more stable in its native form, because apart of weak interactions between R groups, it also presents other stronger interactions, as those including covalent bonds  

Explanation:

For example, covalent bonds between sulfur atoms when disulfide bridges are built.  These links are very difficult to break and maintains the protein shape.  Disulfide bonds are a few but they use to incide in the structure of native proteins

8 0
3 years ago
What is one interaction between the geosphere and the cryosphere
nordsb [41]
The water cycle is one interaction between the geosphere and the cryosphere.
7 0
3 years ago
It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t
irakobra [83]

Answer : The molar mass of the unknown gas will be 79.7 g/mol


Explanation : To solve this question we can use graham's law;


Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;


So here, Rate of gas 2 / Rate of gas 1 = \sqrt{(molar mass 1 / molar mass 2)}

Now, 1.687 = square root [\sqrt{(molar mass 1) / (28.01 g/mol N_{2})} ]


When we square both the sides we get;


2.845 = (molar mass 1) / (28.01 g/mol N2)


On rearranging, we get,


2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

3 0
3 years ago
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