Answer:
Mass of carbon dioxide produced = 52.8 g
Explanation:
Given data:
Mass of carbon react = 14.4 g
Mass of oxygen = 56.5 g
Mass of oxygen left = 18.1 g
Mass of carbon dioxide produced = ?
Solution:
C + O₂ → CO₂
Number of moles of C:
Number of moles = mass/ molar mass
Number of moles = 14.4 g/ 12 g/mol
Number of moles = 1.2 mol
18.1 g of oxygen left it means carbon is limiting reactant.
Now we will compare the moles of C with CO₂.
C : CO₂
1 : 1
1.2 : 1.2
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 1.2 mol × 44 g/mol
Mass = 52.8 g
Answer:
The direct channels by which extraction can affect the economy are of particular interest. The development of mines or oil and gas fields can result in significant shocks to a regional economy, generating jobs and drawing in capital from other regions and countries.
Explanation:
Answer: When you strike a tennis ball with a racket, the slight springyness of the racket allows the ball to propel forward
Explanation:
The value of the Gibbs free energy shows us that the reaction is spontaneous.
<h3>What is the Gibbs free energy?</h3>
The Gibbs free energy is a quantity that helps us to be able to determine the spontaneity of a reaction.
In order to obtain the Gibbs free energy, we must obtain the Ecell as follows; 0.799V - (-0.402V) = 1.201 V
Now;
△G = -nFEcell
△G = -(2 * 96500 * 1.201)
△G = -232kJ/mol
Learn more about free energy:brainly.com/question/15319033?
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Answer:
25 mM Tris HCl and 0.1% w/v SDS
Explanation:
A <em>10X solution</em> is ten times more concentrated than a <em>1X solution</em>. The stock solution is generally more concentrated (10X) and for its use, a dilution is required. Thus, to prepare a buffer 1X from a 10X buffer, you have to perform a dilution in a factor of 10 (1 volume of 10X solution is taken and mixed with 9 volumes of water). In consequence, all the concentrations of the components are diluted 10 times. To calculate the final concentration of each component in the 1X solution, we simply divide the concentration into 10:
(250 mM Tris HCl)/10 = 25 mM Tris HCl
(1.92 M glycine)/10 = 0.192 M glycine
(1% w/v SDS)/10 = 0.1% w/v SDS
Therefore the final concentrations of Tris and SDS are 25 mM and 0.1% w/v, respectively.
