Answer:
MgBr₂ + AgNO₃ => Mg(NO₃)₂ + AgBr
Explanation:
Find the element symbol and charge of each element on the periodic table. For polyatomic ions (nitrate), reference your polyatomic ions chart. Use the "partner's charge" rule to find the number of atoms in each compound.
Charges are written as superscripts. "1" is usually not written, just the + or - sign. The charge of silver is 1, which is the (I) bracket roman numeral 1. It is indicated like that because it is multivalent, meaning it has more than one possible charge.
<u>Write each element as an ion</u> (with the charge).
Magnesium is Mg²⁺
Bromide is Br⁻
Silver(I) is Ag⁺
Nitrate is (NO₃)⁻
<u>Write each compound.</u>
REACTANTS SIDE
Magnesium bromide
Mg²⁺Br⁻ Cross over the partner's charge. Since Br is charge 1, Mg has 1 atom. Since Mg has charge 2, Br has 2 atoms.
MgBr₂
Silver(I) nitrate
Ag⁺(NO₃)⁻
AgNO₃ Both have 1 atom because each partner's charge was 1. You do not need to write brackets if nitrate only has 1 atom.
PRODUCTS SIDE
Magnesium nitrate
Mg²⁺(NO₃)⁻
Mg(NO₃)₂ Nitrate has 2 atoms because magnesium's charge is 2.
Silver(I) bromide
Ag⁺Br⁻
AgBr Both have 1 atom.
Write the compounds into an equation. Reactants go on the left side, products go on the right side. Between the reactants and products, write an arrow.
MgBr₂ + AgNO₃ => Mg(NO₃)₂ + AgBr
Answer:
6.72M of HNO3
Explanation:
In the problem you are diluting the original HNO3 solution by the addition of some water. The final volume is:
290.7mL + 350.0mL = 640.7mL
And you are diluting the solution:
640.7mL / 350.0mL = 1.8306 times
As the original concentration was 12.3M, the final concentration will be:
12.3M / 1.8306 =
<h3>6.72M of HNO3</h3>
Go back and review data, and look for an alternate source. This time, they should check to make sure they are getting their data from a reliable source.
Answer:
Volume of methanol present in solution = less than 3 ml
Explanation:
Given:
Volume of solution = 750 ml
Find:
Volume of methanol present in solution
Computation:
Volume of methanol present in solution = Volume of solution x 0.4%
Volume of methanol present in solution = 750 ml x 0.4%
Volume of methanol present in solution = 3 ml
Volume of methanol present in solution = less than 3 ml