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What is the empirical formula for 13 percent magnesium and 87 percent bromine?
The empirical formula for a compound containing 13% magnesium and 87% bromine is MgBr2. This is because the ratio of magnesium to bromine atoms in the compound is 1:2, which corresponds to the formula MgBr2.
The empirical formula is CH2O. What is the molecular formula of this compound?
The molecular formula of a compound is a multiple of its empirical formula, so the molecular formula is a multiple (in this case, 6 times) of CH2O, giving C6H12O6. This molecular formula corresponds to glucose, a common sugar.
Empirical formula for potassium manganate?
The empirical formula for potassium manganate is KMnO4.
What is the empirical formula for ribose?
The empirical formula for ribose is C5H10O5.
Simplest empirical formula for allicin?
C6H10OS2. Molecular and empirical are the same for Allicin.
What is the empirical formula of a compound that contains 36.1 ca and 63.9 cl?
To find the empirical formula, we first convert the masses of each element to moles. For calcium (Ca), with a molar mass of approximately 40.1 g/mol, 36.1 g of Ca corresponds to about 0.90 moles. For chlorine (Cl), with a molar mass of approximately 35.5 g/mol, 63.9 g of Cl corresponds to about 1.80 moles. The ratio of moles of Ca to Cl is 0.90:1.80, which simplifies to 1:2. Thus, the empirical formula of the compound is CaCl₂.
What is the empirical formula for 13 percent magnesium and 87 percent bromine?
The empirical formula for a compound containing 13% magnesium and 87% bromine is MgBr2. This is because the ratio of magnesium to bromine atoms in the compound is 1:2, which corresponds to the formula MgBr2.
What is the empirical formula of a compound containing 60.0 sulfer and 40.0 oxygen by mass?
To find the empirical formula, first convert the mass percentages to moles. For sulfur (S), 60.0 g corresponds to approximately 1.88 moles (60.0 g / 32.07 g/mol), and for oxygen (O), 40.0 g corresponds to about 2.50 moles (40.0 g / 16.00 g/mol). Next, divide both mole values by the smaller value (1.88), yielding a ratio of approximately 1 S to 1.33 O. Simplifying this ratio gives an empirical formula of ( S_3O_4 ).
The empirical formula is CH2O. What is the molecular formula of this compound?
The molecular formula of a compound is a multiple of its empirical formula, so the molecular formula is a multiple (in this case, 6 times) of CH2O, giving C6H12O6. This molecular formula corresponds to glucose, a common sugar.
Is the empirical formula molar mass less than or equal to the actual molar mass for all compound?
The empirical formula molar mass is the mass of the simplest whole-number ratio of the elements in a compound, while the actual molar mass corresponds to the molar mass of the compound's molecular formula. The empirical formula molar mass is always less than or equal to the actual molar mass because the empirical formula represents the smallest ratio of atoms, which can be multiplied to obtain the molecular formula. Therefore, for compounds with a molecular formula that is a multiple of the empirical formula, the empirical molar mass will be less than the actual molar mass.
What is the probability found by experimenting?
It is experimental or empirical probability.It is experimental or empirical probability.It is experimental or empirical probability.It is experimental or empirical probability.
What is reasoning process?
empirical
What is the translation in Tagalog of empirical?
Tagalog of empirical: impiryo
What is the empirical formula of C10H4?
It is an empirical formula.
Empirical formula for potassium manganate?
The empirical formula for potassium manganate is KMnO4.
What is empirical research?
empirical research or empirical study means: data has already been collected and analyzed.
What is the molecular formula of a substance that has an empirical formula of C2H5 and a molecular mass of 58 grams per mole?
The empirical formula of C2H5 corresponds to an empirical mass of 29 g/mol. To find the molecular formula from the empirical formula and molecular mass, divide the molecular mass by the empirical mass to get the "scaling factor" (58 g/mol ÷ 29 g/mol = 2). Multiply the subscripts in the empirical formula by the scaling factor to get the molecular formula: C2H5 x 2 = C4H10. So, the molecular formula is C4H10.
