1143.4grams of solute are needed to make 2.50L of a 1.75M solution of Ba(NO3)2. Details about molarity can be found below.
<h3>How to calculate mass?</h3>
The mass of a substance can be calculated by multiplying the number of moles by its molar mass.
However, the number of moles can be calculated by using the following formula:
Molarity = no of moles ÷ volume
no of moles = 1.75 × 2.5
no of moles = 4.38mol
molar mass of Ba(NO3)2 = 261.34g/mol
mass of Ba(NO3)2 = 261.34 × 4.38
mass of Ba(NO3)2 = 1143.4grams.
Therefore, 1143.4grams of solute are needed to make 2.50L of a 1.75M solution of Ba(NO3)2.
Learn more about mass at: brainly.com/question/19694949
Answer:
density of a piece of metal = 7 gr/ml
Explanation:
See the file please
Answer:
polar covalent bonds
Explanation:
Phosphorus trichloride is a chemical compound of phosphorus and chlorine, having the chemical formula PCl 3. It has a trigonal pyramidal shape, owing to the lone pairs of electrons on the phosphorus atom.
Chlorine is more elctronegative than phosphorus . Phosphorus has a lone pair of electrons and every chlorine atom has three pairs of electrons . The molecule is not symmetrical as the shared electron pairs are closer to the chlorine atoms than the phosphorus atoms , the chlorine atoms are shifted towards one side of the center of the molecule , thus a net dipole effect is created. This is the reason why the molecule is polar .
Basically since potassium chloride is an ionic compound as it consists of a metal and a nonmetal, the potassium atom will donate one of its valence electrons to chlorine that will accept it and as a result produce oppositely charged ions, where the K + ion and the Cl - ion will attract forming an ionic bond. The compound that results is potassium chloride.
Answer:
because of catenation of carbon.
Explanation:
Catenation is the binding of an element to its self through covalent bonds to form chain or ring molecules. carbon is able to form continuous links with other carbon atoms which is the reason for the existence of a large number of organic compounds.
