Both are caused by human ignorance. Both have the ability to kill those in them. Both can become compact to create new material.
Answer:
- <em>The mystery substance is</em> <u>C. Bromine (Br) </u>
Explanation:
<em>Argon (Ar) </em>is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.
<em>Scandium (Sc) </em>is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.
Both <em>bromine</em> and <em>iodine</em> are halogens (group 17 of the periodic table).
The freezing point of bromine is −7.2 °C, and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.
The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.
Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.
You can find in the internet that bromine vapour over hot iron reacts producing iron(III) bromide. Also, that bromine vapors are red-brown.
Therefore, <em>the mystery substance is bromine (Br).</em>
When transferring flammable liquids from storage drums to smaller electrically conductive containers, bonding and grounding are required. In any workplace, you should do the same whenever you move these liquids between conductive containers.
<h3>What exactly occurs in this case?</h3>
Ground dispensing drums in the area for storing and dispensing flammable liquids. Connecting the container to an already grounded, electrically conductive object grounds it. This could be a grounded metal construction framework, a buried metal plate, a metallic subsurface gas piping system, or metal water pipes. Sparks are prevented from discharging by bonding the two containers together and grounding one of them. All connections for bonding and grounding must be made from bare metal to bare metal.
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Answer:
20.76 L OF CO2 WILL BE PRODUCED BY 45 G OF METHANE.
Explanation:
Equation of the reaction:
CH4 + 02 --------> CO2 + 2H20
Molar mass of methane = ( 12+ 1*4) g/mol = 16 g/mol
Calculate the number of moles present in 45 g of methane
1 mole of methane = 16 g / mol of methane
(45 / 16) mole of methane = 45 g of methane
= 2.8125 moles
Using the ideal gas equation:
PV = nRT
P = 1 atm
n = 2.812 moles
T = 90 C
R = 0.082 L atm/ mol C
V = unknown
So we have:
V = nRT / P
V = 2.8125 * 0.082 * 90 / 1
V = 20.756 L
In the production of CO2 by 45 g of methane, 20.756 L of methane was used.
Then, the volume of CO2 produced by this volume will be 20.756 L since 1 mole of methane produces 1 mole of CO2.
In other words;
1 mole of CH4 = 1 mole of CO2
22.4 dm3 of CH4 = 22.4 dm3 of CO2
20.76 DM3 = 20.76 dm3
The volume of CO2 produced will therefore be 20.76 L