Answer:
=> 1366.120 g/mL.
Explanation:
To determine the formula to use in solving such a problem, you have to consider what you have been given.
We have;
mass (m) = 25 Kg
Volume (v) = 18.3 mL.
From our question, we are to determine the density (rho) of the rock.
The formula:
First let's convert 25 Kg to g;
1 Kg = 1000 g
25 Kg = ?
= 25000 g
Substitute the values into the formula:
= 1366.120 g/mL.
Therefore, the density (rho) of the rock is 1366.120 g/mL.
Molar mass is the mass of 1 mol of substance.
Molar masses of compounds can be calculated by the sum of the products of molar masses of individual atoms by number of corresponding individual atoms.
Compound formula is C₉H₈O₄
the molar masses of the atoms making up the compound
C - 12 g/mol x 9 C = 108
H - 1 g/mol x 8 H = 8
O - 16 g/mol x 4 O = 64
therefore molar mass of aspirin = 108 + 8 + 64 = 180 g/mol
answer is 3.180
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:
The symbol of each element is, Ne, Na, Mg, and Al.
Explanation:
Below is the list of elements that has an atomic mass of less than 19.3 u.
The atomic mass of Neon is 20.1797 u and the atomic number is 10.
The atomic mass of Sodium is 22.989769 u and the atomic number is 11.
The atomic mass of Magnesium is 24.305 u and the atomic number is 12.
The atomic mass of Aluminium is 26.981539 u and the atomic number is 13.
Here, the symbol of each element is, Ne, Na, Mg, and Al.
