From his experiment j.j. Thomson concluded that

A sample of sugar (C12H22O11) contains

Mariana [72]

Answer:

0.25 mol

Explanation:

Use the formula n=N/NA

n= number of mols

N = number of particles

Nᵃ = Avogadros constant = 6.02 x $\\10^{23$

So, n= $\frac{1.505 X 10^{23} }{6.02 X 10^{23}}$

The 10 to the power of 23 cancels out and you are left with 1.505/6.02, which is approximately 1/4. This is the same as 0.25 mol.

Hope this helped :)

3 0

3 years ago

A Sattelite is a (n)

uysha [10]

Satellite, also called artificial satellite, is a computer-controlled machine or objects that is launched into the space by a rocket, called the launch vehicle and placed in the orbit of a celestial object.

6 0

4 years ago

Based on the greenhouse effect, if the amount of carbon dioxide in the air decreased, what would happen?

choli [55]

A) the average global temp. Would decrease

7 0

4 years ago

Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −

irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

$H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}$

The chemical equation for the second dissociation reaction follows:

$H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}$

The chemical equation for the third dissociation reaction follows:

$HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}$

To form a buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a$ of second dissociation process

To calculate the $pK_a$ , we use the equation:

$pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21$

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})$

$pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})$

We are given:

$pK_a2$ = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

$[HPO_4^{2-}]$ = concentration of conjugate base

$[H_2PO_4^{-}]$ = concentration of weak acid

Hence, to calculate the pH of the buffer composed of $H_2PO_4^-\text{ and }HPO_4^{-2}$ , we use the $K_a2$

3 0

4 years ago

25. What is the equilibrium constant expression for the following? 2Hg (g) + O₂(g) → 2HgO (s) A. K= [Hg] [0₂]. B. K= [HgO]/([Hg]

boyakko [2]

Answer:

2 Hg (g) + o2 ---> 2 H2o

equilibrium constant

K = (c) (d) / (a)(b)

K= ( H2o)²/(He) ²(o2)

5 0

2 years ago