by pencil or pen
It depends on 0.5 measured in what units. The number of 0.5 grams in 2 mg will be different from the number of 0.5 nanograms, for example.
F=mg where g is the gravitational constant, and the independent variable in force calculations
Divide the number of milligrams be 1000 to convert to grams: 615 mg = 0.615 g
If N is the number, and f is the number that you want to test as a possible factor, then first of all:test N > f (this must be true, the factors are always smaller in magnitude)next perform N ÷ f (N divided by f). If the quotient (answer to a division problem) is a whole number with no remainder or fractional part, then f is a factor of N.If the quotient is not a whole number (meaning there is a remainder), then f is not a factor.
To find the number of fluorine (F) atoms in 5.88 mg of ClF3, first calculate the molar mass of ClF3. ClF3 has a molar mass of 83.45 g/mol. Convert 5.88 mg to grams (0.00588 g) and then use the molar mass to find the number of moles of ClF3. Since there are 3 F atoms in each molecule of ClF3, multiply the number of moles by Avogadro's number (6.022 x 10^23) and then by 3 to find the number of F atoms. In this case, there are approximately 4.24 x 10^20 F atoms in 5.88 mg of ClF3.
MG F-type was created in 1931.
Yes!
Ionic bond: Mg2+ + 2 F- --> MgF2
The atomic number of Mg (Magnesium) is 12.
Magnesium (Mg) and fluorine (F) would form magnesium fluoride (MgF2) through ionic bonding.
yes, Thomas Park D'Invilliers
Katherine Anne Porter
F = mg
Mg and F is most likely to form an ionic compound because magnesium (Mg) has a +2 charge and fluorine (F) has a -1 charge, leading to a strong attraction between the two. Oxygen (O) and chlorine (Cl) also form ionic compounds, but the attraction between Mg and F is stronger due to their larger difference in electronegativity.
The oxidation number of Mg in Mg3N2 is +2. Each N has an oxidation number of -3, so for the compound to be neutral, the magnesium atoms must have an oxidation number of +2.
The PEN (Product Environmental Profile) number for calcium would vary depending on the specific form or compound of calcium being referred to. It is important to specify the exact form of calcium to determine its corresponding PEN number.