1 wavelength, 2 crest, 3 trough, 4 wave height <3
Answer:
There are 3, 64 moles of NaCl.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate themoles in 213 grams of NaCl, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl = 23 g + 35, 5 g= 58, 5 g/ mol
58,5 g ------1 mol NaCl
213 g---------x= (213 g x 1 mol NaCl)/ 58, 5 g= <em>3, 64 mol NaCl</em>
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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Answer is: ph value of pyridine solution is 9.1.
Chemical
reaction: C₅H₅N +
H₂O → C₅H₅NH⁺ + OH⁻.<span>
c(pyridine - C</span>₅H₅N)
= 0.115M.<span>
Kb(C</span>₅H₅N)
= 1.4·10⁻⁹.
[C₅H₅NH⁺] = [OH⁻] = x; equilibrium concentration.<span>
[</span>C₅H₅N] =
0.115 M - x.
Kb = [C₅H₅NH⁺] · [OH⁻] / [C₅H₅N].
1.4·10⁻⁹ = x² / (0.115 M -x)
Solve quadratic equation: x = [OH⁻] = 0.0000127 M.<span>
pOH = -log(0.0000127 M) = 4.9</span>
<span>pH = 14 - 4.9 = 9.1.</span>