Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Answer:
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Explanation:
Gold does not tarnish, rust or corrode. Due to its wonderful qualities and its luster, gold is considered the most important metal in jewellery making. As pure gold is too soft for everyday wear, it is alloyed with a mixture of metals in order to make the gold harder, so it can be used for jewellery.
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Answer:
a. 58.5 g/mol
b. 0.1 mol
Explanation:
a.
The molar mass of Na is 23.0 g/mol. The molar mass of Cl is 35.5 g/mol. The molar mass of NaCl is:
M(Na) + M(Cl) = 23.0 g/mol + 35.5 g/mol = 58.5 g/mol
b. A healthy adult should eat no more than 6 g of salt in one day. The moles corresponding to 6 g of NaCl are:
6 g × (1 mol/58.5 g) = 0.1 mol
Answer:
Explanation:
1) A fulcrum is a pivot point that plays a central role (not necessarily located at the center) in a lever. The fulcrum of the attached picture has been circled (in blue).
2) The object placed on this lever's measurement tray is balanced by placing it at the center of the tray. This is the standard way of placing objects on any balance.
