Help with number 20-22 please

Label parts of the cell

myrzilka [38]

Is this what you mean? If not I’ll try to help you

4 0

3 years ago

Pllllllsssssssssss helpppp Determine the Mass Number of the Following Elements:

aleksklad [387]

Al- 28
B- 11
C- 14
Sc- 46
Na- 24
Mass number= proton+neutrons

5 0

3 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

3 years ago

A Canadian dime weighs 1.75g and has a thickness of 1.22mm. A Canadian quarter weighs 4.4g and has a thickness of 1.58mm. You ha

balu736 [363]

The mass of the total number of coins is 1405 g.

Since the thickness of each dime is 1.22 mm and we have each pile being 61.0 mm tall, we need to determine the number of dimes in each pile by dividing the height of the pile by the thickness of each dime.

so, number of dimes in each pile = height of each pile/thickness of each dime = 61.0 mm/1.22 mm = 50 dimes.

Since we have 6 piles, the total number of dimes in all the piles is 6 piles × 50 dimes/pile = 300 dimes.

So, there are 300 dimes in the 6 piles.

Since each dime weighs 1.75 g, 300 dimes will weigh 1.75 g/dime × 300 dimes = 525 g.

In the pile of quarters, we have $ 50 worth of quarters.

Since each quarter = $ 0.25, the number of quarters in that pile is total worth of quarter/value of one quarter = $ 50/$ 0.25 = 200 quarters.

Since each quarter weighs 4.4 g, 200 quarters will weigh 4.4 g/quarter × 200 quarters = 880 g

So, the mass of the total number of coins = total mass of dimes + total mass of quarters = 525 g + 880 g = 1405 g

So, the mass of the total number of coins is 1405 g.

Learn more about mass of coin here:

brainly.com/question/19194652

5 0

3 years ago

What is the oxidation number of chromium in the ionic compound ammonium dichromate (nh4)2cr2o7?

lesya692 [45]

The oxidation number of an element is the number of electrons that are gained or lost by the element to form a chemical bond.
the net oxidation number of the ion is its charge.
compound (NH₄)₂Cr₂O₇ is made of cation - NH₄⁺ and anion - Cr₂O₇²⁻
oxidation number of ion - -2
(oxidation number of Cr x 2 Cr atoms) +(oxidation number of O x 7 O atoms )= -2
oxidation number of Cr = y
oxidation number of O = -2
(y x 2) + (-2 x 7) = -2
2y - 14 = -2
2y = 12
y = + 6
therefore oxidation number of Cr is +6

4 0

4 years ago