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swat32
3 years ago
13

2 C4H10 + 13 O2--> 8 CO2 + 10 H2O

Chemistry
1 answer:
AleksAgata [21]3 years ago
7 0

Answer:

4.14 x 10²⁴ molecules CO₂

Explanation:

2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O

To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.

4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀

100 grams C₄H₁₀          1 mol C₄H₁₀             8 mol CO₂          
--------------------------  x  ----------------------  x  ---------------------  
                                        58.124 g              2 mol C₄H₁₀          

    6.022 x 10²³ molecules
x  ------------------------------------  =  4.14 x 10²⁴ molecules CO₂
              1 mol CO₂

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Answer:

Yes

Explanation:

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The given equation is balanced. It shows the emission of alpha particle which is helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4  and atomic number less than 2 as compared to parent atom the starting atom.

Properties of alpha radiation:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei.

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

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7 0
3 years ago
A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi
attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}

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At equilibrium, E = 0, therefore:

E^{o}  = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] =  log[ox] -  \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] -  \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 -  \frac{2x5.45x10^{-2}  }{0.0591}}\\\\

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4 years ago
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In electron capture , the atom have only one less which is daughter isotopes than electron than atomic number of parent isotopes.

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