<span>All metals have similar properties BUT, there can be wide variations in melting point, boiling point, density, electrical conductivity and physical strength.<span>To explain the physical properties of metals like iron or sodium we need a more sophisticated picture than a simple particle model of atoms all lined up in close packed rows and layers, though this picture is correctly described as another example of a giant lattice held together by metallic bonding.</span><span>A giant metallic lattice – the <span>crystal lattice of metals consists of ions (NOT atoms) </span>surrounded by a 'sea of electrons' that form the giant lattice (2D diagram above right).</span><span>The outer electrons (–) from the original metal atoms are free to move around between the positive metal ions formed (+).</span><span>These 'free' or 'delocalised' electrons from the outer shell of the metal atoms are the 'electronic glue' holding the particles together.</span><span>There is a strong electrical force of attraction between these <span>free electrons </span>(mobile electrons or 'sea' of delocalised electrons)<span> (–)</span> and the 'immobile' positive metal ions (+) that form the giant lattice and this is the metallic bond. The attractive force acts in all directions.</span><span>Metallic bonding is not directional like covalent bonding, it is like ionic bonding in the sense that the force of attraction between the positive metal ions and the mobile electrons acts in every direction about the fixed (immobile) metal ions of the metal crystal lattice, but in ionic lattices none of the ions are mobile. a big difference between a metal bond and an ionic bond.</span><span>Metals can become weakened when repeatedly stressed and strained.<span><span>This can lead to faults developing in the metal structure called 'metal fatigue' or 'stress fractures'.</span><span>If the metal fatigue is significant it can lead to the collapse of a metal structure.</span></span></span></span>
Answer:
this is fairly simple if you have a periodic table with you.
Explanation:
atomic number 17 is Cl mass numer is 35.45 for protons neutron and electron you can just look that up on google. atomic number is where it is at on the periodic table and the mass number is in the little square at the bottom.
The heat required to vaporize 43.9 g of acetone at its boiling point is calculated as below
the heat of vaporization of acetone at its boiling point is 29.1 kj/mole
find the moles of acetone = mass/molar mass
= 43.9g /58 g/mol =0.757 moles
heat (Q) = moles x heat of vaporization
= 29.1 kj/mole x 0.757 moles = 22.03 kj
274 mL H3 O+ and fully neutralized
It will take one teaspoon of Mg(OH)2 to completely neutralize 2.00×10^2mL of H3O+.
<h3>What is the purpose of milk of magnesia?</h3>
- For a brief period of time, this medicine is used to relieve sporadic constipation.
- It is an osmotic laxative, which means that it works by drawing water into the intestines, which aids in causing bowel movement.
<h3>What dosage of milk of magnesia is recommended for constipation?</h3>
- Take Milk of Magnesia once day, preferably before bed, in divided doses, or as prescribed by a physician.
- suggested dosage: 30 mL to 60 mL for adults and kids 12 years of age and older. 15 mL to 30 mL for children aged 6 to 11 years.
learn more about milk of magnesia here
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the question you are looking for is
People often take milk of magnesia to reduce the discomfort associated with acid stomach or heartburn. The recommended dose is 1 teaspoon, which contains 4.00x 10^{2} mg of Mg(OH)_2. What volume of an HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia? If the stomach contains 2.00x10^{2}mL of pH 1.3 solution, is all the acid neutralized? If not, what fraction is neutralized?
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
