Answer:
47.9L of oxygen would be required
Explanation:
<em>Assuming oxygen gas is at STP</em>
Based on the reaction:
S₈ + 12 O₂ → 8SO₃
1 mole of S₈ (Molar mass: 256.52g/mol) reacts with 12 moles of oxygen.
To solve this question we must find the moles of S₈ added. With the reaction we can find the moles of O₂ and using PV = nRT we can find the volume of oxygen required:
<em>Moles S₈:</em>
45.7g * (1mol / 256.52g) = 0.178 moles S₈
<em>Moles O₂:</em>
0.178 moles S₈ * (12 moles O₂ / 1mol S₈) = 2.138 moles O₂
<em>Volume O₂:</em>
PV = nRT
V = nRT / P
<em>Where V is volume</em>
<em>n are moles = 2.138 moles</em>
<em>R is gas constant 0.082atmL/molK</em>
<em>T is absolute temperature = 273.15K at sTP</em>
<em>P is pressure = 1 atm at STP</em>
<em />
V = 2.138mol*0.082atmL/molK*273.15K / 1atm
V = 47.9L of oxygen would be required
Because we are moving at a constant rate (seeing as we're on earth and the earth is spinning) but the moon does not spin as it revolves around us, therefore we only see the one side of it.
13% of total energy ( net energy ) is obtained from renewable sources and about 46% of it comes from hydroelectric power plants.
Answer:
Potassium (K) and Fluorine (F)
Explanation:
A salt is formed with a metal and non-metal element from the periodic table of elements. if you look at the online ptable.com the upper right hand corner of each element shows the valence electrons for each element, and how many are in each shell for that element. Fluorine (F) has 7 valence electrons in its outer most shell, which means there is room for 1 more electron since the second shell can hold a max of 8. Potassium(K) has 1 electron in its outer most shell- which means is can fill in the 1 space available that fluorine has in its outer most shell. Since Potassium(K) is a metal and Fluorine(F) is a non-metal they can form an ionic compound, salt.
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.
