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o-na [289]
3 years ago
6

How many miles of co2 are produced if 5.0 moles of c10h22 react with an excess of o2?

Chemistry
1 answer:
Naily [24]3 years ago
3 0
The balanced equation for the above reaction is as follows;
2C₁₀H₂₂ + 31O₂ ---> 20CO₂  + 22H₂O
stoichiometry of C₁₀H₂₂ to CO₂ is 2:20
this means that for every 2 mol of C₁₀H₂₂ that reacts - 20 mol of CO₂ is formed 
therefore when 5.0 mol of C₁₀H₂₂ reacts - 20/2 x 5.0 = 50 mol of CO₂ is formed 
50 mol of CO₂ is produced.
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Calculate the oxidation number of the iodine (I) in each compound: HIO4 = I2 = NaI = HIO3 =
Makovka662 [10]
1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.

2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.

3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.

4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.
8 0
3 years ago
Read 2 more answers
What is the mass of 3.00 moles of magnesium chloride, MgCl2? Express your answer with the appropriate units.
xenn [34]
<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows: Mg=24 g/mole Cl=38 g/mole Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so: (38 g/mole*2 moles)+24 g/mole=100 g/mole Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows: 3.00 moles * 100 g/mole = 300 g</span>
5 0
3 years ago
Limestone stalactites and stalagmites are formed in caves by the following reaction: Ca2 (aq) 2HCO−3(aq)→CaCO3(s) CO2(g) H2O(l)
Vesna [10]

Answer : The value of \Delta E for this reaction is 36.18 kJ

Explanation :

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

\Delta E=q+w

where,

\Delta E = internal energy  of the system

q = heat added or rejected by the system

w = work done

As we are given that:

q = 38.65 kJ

w = -2.47 kJ (system work done on surrounding)

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

\Delta E=38.65kJ+(-2.47kJ)

\Delta E=36.18kJ

Therefore, the value of \Delta E for this reaction is 36.18 kJ

6 0
3 years ago
What is the longitude of the international date line
sveta [45]

Answer:

The International Date Line passes through the mid-Pacific Ocean and roughly follows a 180 degrees longitude north-south 

line on the Earth. It is located halfway round the world from the prime meridian—the zero degrees longitude established in Greenwich

8 0
3 years ago
The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
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