Answer:

Explanation:
Hello,
In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

Therefore, the empirical formula is:

Best regards.
Hey there!
Molar mass N2 = 28.01 g/mol
Therefore:
28.01 g N2 -------------- 6.02*10²² molecules N2
( mass N2 ?? ) ----------- 25,000 molecules N2
mass N2 = ( 25,000 * 28.01 ) / ( 6.02*10²³ )
mass N2 = 700250 / 6.02*10²³
mass N2 = 1.163*10⁻¹⁸ g
Hope that helps!
Answer : Option 3) Wave/Particle duality.
Explanation : The experiment on discovery of photoelectric effect revealed about the photoelectrons of light that can behave as particle or waves.
The photoelectric effect is observed when the emission of electrons or other free carriers occurs on shining a light radiation on a material. The electrons emitted from this can be called photo electrons. These photoelectrons may behave as wave or particle in duality which holds that light and matter exhibit properties of both waves and of particles.
Sucrose is a non ionic compound. It does liberates ion when dissolved in water unlike NaCl or other salts which dissolve in water and produce respective cations and anions.
Thus if any amount of sucrose is dissolved in water, it will form non ionic aqueous solution (it will dissolve completely). Thus sucrose solution being non electrolytic will not conduct electricity in aqueous solution.
the bulb will not light up as sucrose will remain in molecular form only
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)

[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:

![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)

=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96