Answer:
Explanation:
Na react with H2O to form NAOH
2 Na+2H2O....................2NAOH + H2
Ca react with water and form calcium hydroxide
Ca + 2H2O........................Ca(OH)2
Mg react with water and form Magnesium hydroxide
Mg +2H2O .........................Mg(OH)2 however this coating of mg(oh)2 prevent it from further reaction
Fe react with water and form ferric hydride
3Fe +H2O.......................2 FeH +FeO
copper do not react with water
To determine the amount of a substance in units of moles from units of grams, we need to determine the molar mass of the substance. <span>The </span>molar mass<span> is the </span>mass<span> of a given chemical element or chemical compound (g) divided by the amount of substance (mol). For CuF2, the molar mass </span><span>101.543 g/mol. We calculate as follows:
100.0 g CuF2 ( 1 mol / 101.543 g) = 0.98 mol CuF2</span>
Effect of Two-Step Homogenization on the Evolution of Al3Zr Dispersoids in Al-0.3Mg-0.4Si-0.2Zr Alloy Al3Zr nano-particles can be introduced in Al-Mg-Si 6xxx alloys to improve their elevated temperature behavior and recrystallization resistance. The effect of two-step homogenization treatments on
the precipitation of Al3Zr dispersoids in Al-0.3Mg-0.4Si-0.2Zr alloy was investigated and compared to
<h3>What is
Homogenization?</h3>
Any of a number of methods, including homogenization and homogenisation, are used to uniformly combine two liquids that are insoluble in one another. To do this, one of the liquids is changed into a state in which very minute particles are evenly dispersed across the other liquid. The process of homogenizing milk, in which the milk fat globules are equally distributed throughout the remaining milk and reduced in size, is a classic example. In order to create an emulsion, two immiscible liquids (i.e., liquids that are not soluble in all amounts one in another) must be homogenized (from "homogeneous"; Greek, homos, same + genos, kind)[2] (Mixture of two or more liquids that are generally immiscible).
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Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)