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Ksivusya [100]
3 years ago
8

La massa d'una molècula de glucosa (C6H12O6) és:

Chemistry
1 answer:
ch4aika [34]3 years ago
8 0
The molecular mass of glucose is
6*(12.011) + 12*(1.0080) + 6(15.999) = 180.156 g/mol
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Anna35 [415]

Answer:

5 1 2 4and 3 this is correct way

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2 years ago
7 When carbon is heated in a limited supply
irakobra [83]

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5 0
3 years ago
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Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas.Express your a
ioda

Answer:

Explanation:

From the statement of the problem,

  B₂S₃_{s} + H₂O_{l}  →  H₃BO₃_{aq} + H₂S_{g}

              B₂S₃ + H₂O  →  H₃BO₃ + H₂S

We that the above expression does not conform with the law of conservation of mass:

To obey the law, we need to derive a balanced reaction equation:

   Let us use the mathematical method to obtain a balanced equation.

let the balanced equation be:

                        aB₂S₃ + bH₂O  →  cH₃BO₃ + dH₂S

where a, b, c and d will make the equation balanced.

  Conservating B: 2a = c

                          S: 3a = d

                          H: 2b = 3c + 2d

                           O: b = 3c

   if a = 1,

      c = 2,

      b = 6,

      2d = 2(6) - 3(2) = 6, d = 3

Now we can input this into our equation:

                     B₂S₃ + 6H₂O  →  2H₃BO₃ + 3H₂S

    B₂S₃_{s} + 6H₂O_{l}  →  2H₃BO₃_{aq} + 3H₂S_{g}

4 0
3 years ago
Isotope b has a half-life of 3 days. a scientist measures out 100 grams of this substance. after 6 days has passed, the scientis
Brilliant_brown [7]
So half life is the time taken for a sample to decay to half its original mass, its a constant and applies to any original mass, it could be 5g or 1kg, it will take the same amount of time for the original mass to half. In this case the half life is 3 days.

After 3 days the sample will be at half its original mass, now 50g. 

Now we can treat the 50g as if its a new sample. After another 3 days (6 days in total) there will be half of 50g left, = 25g. 


6 0
3 years ago
The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s
raketka [301]

Answer : The equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

Explanation :

The dissociation of acid reaction is:

                       C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-

Initial conc.        c                                 0                0

At eqm.             c-x                                 x                x

Given:

c = 7.0\times 10^{-2}M

K_a=6.3\times 10^{-5}

The expression of dissociation constant of acid is:

K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}

K_a=\frac{(x)\times (x)}{(c-x)}

Now put all the given values in this expression, we get:

6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}

x=2.1\times 10^{-3}M

Thus, the equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

4 0
2 years ago
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