Answer:
All of the above.
Explanation:
Hello there!
In this case, according to the definition of the Lewis structures, which are represented by the valence electrons, we first identify that the N atom has five valence electrons and each fluorine has seven valence electrons.
In such a way, we cans say that N is the central atom due to its lower electronegativity, the molecule has 7+7+7+5=26 valence electrons and the three F-N bonds are covalent, therefore the answer is all of the above.
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Answer:
Explanation:
1. Cu(NO3)2 + 2NaOH → Cu(OH)2 + 2NaNO3
2. FeSO4 + 2 NaOH = Fe(OH)2 + Na2SO4
3. Fe(NO3)3 + 3 NaOH = Fe(OH)3 + 3 NaNO3
For the second part of 3 since it tells us NaOH is excess we dont care about it. We use our balanced equation 3 to get the correct molar ratios to solve this.
1.5mol Fe(NO3)3 x 3mol NaNO3/1mol Fe(NO3)3 x 84.99gNaNO3/1molNaNO3
= 382.455g NaNO3
Answer:
3.5 mL
Explanation:
Let us represent lauric acid with the symbol LaCOOH and the anion as LaCOO-. The reaction between the anion and H2SO4 is;
LaCOO-(aq) + H2SO4(aq) -----> LaCOOH(aq) + HSO4-(aq)
Number of moles of LaCOO- = 2.75 g/200 g/mol = 0.014 moles
Since the mole ratio of the reaction is 1:1, then the amount of H2SO4 required is also 0.014 moles
Then;
n = CV
n = number of moles
C= concentration
V = volume
V = n/C
V = 0.014 moles/4.0 M
V= 3.5 * 10^-3 L
V = 3.5 mL
3.08*10^23 ions of Na+ and Cl- will be present in 30g of NaCl.
Weight of NaCl is equal to sum of weight of one mole of sodium and one mole of chloride.
Mathematically, it can be represented as -
1 mol (Na) + 1 mol (Cl)
= 23.0+ 35.5
= 58.5 g/mol
In 58.5 gram of NaCl, there is one mole of Na or 6.02*10^23 ions
In one gram of NaCl, there will be
=6.02*10^23/58.5 sodium ions
In 30 grams of NaCl, the number of sodium ions will be -
(6.02*10^23*30)/ 58.5
=3.08*10^23 ions
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