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Zepler [3.9K]
3 years ago
7

How many moles of carbon dioxide, co2, are produced if 5.87 moles of glucose, c6h12o6, react?

Chemistry
1 answer:
almond37 [142]3 years ago
3 0
The balanced equation for the reaction is as follows
C₆H₁₂O₆ + 6O₃ --> 6CO₂ + 6H₂O
stoichiometry of glucose to CO₂ is 1:6
when 1 mol of glucose reacts - 6 mol of CO₂ are formed 
therefore when 5.87 mol of glucose reacts - 6 x 5.87 mol = 35.22 mol
therefore 35.2 mol of CO₂ is formed 
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Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq )
Oxana [17]

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

=9

so pH=14-9

=5

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

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3 years ago
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Are protons unstable​
kiruha [24]
If you're including protons that are bound as part of an atomic nucleus, then, yes, under certain conditions.
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3 years ago
Which Scientists’ observations eventually led to the theory of continental drift?
Vitek1552 [10]

Alfred Wegener

Explanation:

Alfred Wegener in 1912, proposed the theory of continental drift from his careful studies of different land masses. From his observations, he suggested the following:

  • Some fossils of tropical plants were found in polar and temperate regions.
  • Fossils of mesosaurus were found in South America and South Africa now separated by the Atlantic ocean.
  • He fitted the continents into a giant land mass puzzles.
  • Coal beds in Europe were similar to those in United States.
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His theory lacked a mechanism which facilitated the plate movement.

Learn more:

Wegener brainly.com/question/5002949

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8 0
3 years ago
I need help on both a and b of question 1
marishachu [46]

Answer:

(a) -0.00017 M/s;

(b) 0.00034 M/s

Explanation:

(a) Rate of a reaction is defined as change in molarity in a unit time, that is:

r = \frac{\Delta c}{\Delta t}

Given the following reaction:

2 N_2O_5 (g)\rightleftharpoons 4 NO_2 (g) + O_2 (g)

We may write the rate expression in terms of reactants firstly. Since reactants are decreasing in molarity, we're adding a negative sign. Similarly, if we wish to look at the overall reaction rate, we need to divide by stoichiometric coefficients:

r = -\frac{\Delta [N_2O_5]}{2 \Delta t}

Reaction rate is also equal to the rate of formation of products divided by their coefficients:

r = \frac{\Delta [NO_2]}{4 \Delta t} = \frac{\Delta [O_2]}{\Delta t}

Let's find the rate of disappearance of the reactant firstly. This would be found dividing the change in molarity by the change in time:

r_{N_2O_5} = \frac{0.066 M - 0.100 M}{200.00 s - 0.00 s} = -0.00017 M/s

(b) Using the relationship derived previously, we know that:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

Rate of appearance of nitrogen dioxide is given by:

r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t}

Which is obtained from the equation:

-\frac{\Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{4 \Delta t}

If we multiply both sides by 4, that is:

-\frac{4 \Delta [N_2O_5]}{2 \Delta t} = \frac{\Delta [NO_2]}{\Delta t}

This yields:

[tex]r_{NO_2} = \frac{\Delta [NO_2]}{\Delta t} = -2\frac{\Delta [N_2O_5]}{ \Delta t} = -2\cdot (-0.00017 M/s) = 0.00034 M/s[tex]

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2 years ago
How many orbitals are in the first energy level? A. 3 B.4 C.2 D.1
WINSTONCH [101]

Answer:

3 orbitals

Explanation:

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