After balancing the equation, what would be the coefficient for water? __NH3+__O2=__NO+H2O
1 answer:
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Answer:
Take a look at the attachment below
Explanation:
Take a look at the periodic table. As you can see, Rubidium is the closest element to Cesium, and happens to have the closest boiling point to Cesium, with only a difference of about 30 degrees.
Respectively, you would think that fluorine should have the least similarity to Cesium with respect to it's boiling point, considering it is the farthest away from the element out of the 4 given. This is not an actual rule, there are no fixed trends of boiling points in the periodic table, there are some but overall the trends vary. However in this case fluorine does have the least similarity to Cesium with respect to it's boiling point, a difference of about 1,546.6 degrees.
<em>Hope that helps!</em>
Answer:
<em><u>The molarity of the CoBr2•4H2O solution is 7.64 × 10-2 M</u></em>
Explanation:
Cobalt (II) bromide tetrahydrate
• Cobalt - A transition metal with Roman numeral (II) → charge: +2 → Co2+
• Bromide - anion from group 7A → -1 charge → symbol: Br-
• Tetrahydrate- tetra- means 4 and hydrate is H2O
The chemical formula of the compound is: CoBr2•4H2O
We then need to determine the number of moles of CoBr2•4H2O since this is the only information missing for us to find molarity. Notice that the volume of the solution is already given.
We’re given the mass of CoBr2•4H2O. We can use the molar mass of CoBr2•4H2O4 to find the moles.
•The molar mass of CoBr2•4H2O is:
CoBr2•4H2O
1 Co x 58.93 g/mol Co = 58.93 g/mol
2 Br x 79.90 g/mol Br = 159.80 g/mol
8 H x 1.008 g/mol H = 8.064 g/mol
4 O x 16.00 g/mol O = 64.00 g/mol
________________________________________
Sum = <u>290.79 g/ mo</u>
The moles of CoBr2•4H2O is:
= 10.0 g CoBr2•4H2O x
= <u>0.0344 mol CoBr2•4H</u>
We know that the volume of the solution is 450 mL.
We can now calculate for molarity:
Convert mL to L → 1 mL = 10-3 L
Formula:
Molarity (M)= Mole of solute / Liters of solution
= 0.0344 mol CoBr2•4H / 450 mL x 1 ml / 10^ -3 L
= 0.0764
= 7.64 × 10-2 mol/L