<h3>
Answer:</h3>
19.3 g/cm³
<h3>
Explanation:</h3>
Density of a substance refers to the mass of the substance per unit volume.
Therefore, Density = Mass ÷ Volume
In this case, we are given;
Mass of the gold bar = 193.0 g
Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm
We are required to get the density of the gold bar
Step 1: Volume of the gold bar
Volume is given by, Length × width × height
Volume = 0.50 cm × 10.0 cm × 2.0 cm
= 10 cm³
Step 2: Density of the gold bar
Density = Mass ÷ volume
Density of the gold bar = 193.0 g ÷ 10 cm³
= 19.3 g/cm³
Thus, the density of the gold bar is 19.3 g/cm³
<h3><u>Answer</u>;</h3>
Groups 14 and 15 each contain metals, nonmetals, and metalloids while Group 13 contains metals and a metalloid, and Group 16 contains metalloids and nonmetals.
<h3><u>Explanation;</u></h3>
- Groups 13–16 of the periodic table contain one or more metalloids, in addition to metals, nonmetals, or both.
- Unlike other groups of the periodic table, which contain elements in one class, groups 13–16 referred to as mixed groups contain elements in at least two different classes. In addition to metalloids, they also contain metals, nonmetals, or both.
- <em><u>Group 14 also known as the carbon group contains carbon which is a non metal, silicon and germanium which are metalloids and tin and lead which are metals.</u></em>
- <em><u>Group 15 also known as the Nitrogen group contains non metals such as oxygen, metalloid tellurium and a metal polonium.</u></em>
Answer:
52 amu
Explanation:
To get the relative atomic mass of the element, we need to take into consideration, the atomic masses of the different isotopes and their relative abundances. We simply multiply the percentages with the masses. This can be obtained as follows:
[89/100 * 52] + [8/100 * 49] + [3/100 * 50]
46.28 + 3.92 + 1.5 =51.7 amu
The approximate atomic mass of element x is 52 amu
If we write the equation of the reaction that will take place, it is:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂CO3
The molar ratio of 2HNO₃ : Na₂CO₃ = 1 : 2
Therefore, we can set up the equation:
M₁V₁ = 2M₂V₂
Where the left side of the equation has the molarity and volume of HNO₃ and the right side has the molarity and concentration of Na₂CO₃. Substituting:
M₁ = (2 x 0.108 x 35.7) / 25
M₁ = 0.308 M
