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serious [3.7K]
3 years ago
9

I'm doing a adopt a element project and I need a slogan for bismuth please tell me a good one

Chemistry
1 answer:
trapecia [35]3 years ago
6 0

Bismuth, it’s elemental!

Bismuth, Agent 83

Bismuth, what a catalyst?

I like you more than Bismuth

Bismuth, we love you

Get things done with Bismuth

It’s a Bismuth thing, you wouldn’t understand

Bismuth, it’s a brittle thing

Got Bismuth?

Bi Bro

Bismuth born Brittle

Silvery by default like Bismuth

Bi the one with Bismuth


http://www.bestslogans.com/list-ideas-taglines/periodic-table-bismuth-slogans/


https://sloganshub.org/bismuth-slogans/

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A solution of 0.0470 M HCl is used to titrate 26.0 mL of an ammonia solution of unknown concentration. The equivalence point is
DanielleElmas [232]

The pH at equivalence point is 12.46

At equivalence point, number of moles of acid, n equals number of moles of base, n'

So, n = n'

CV = C'V' where

  • C = concentration of acid (HCl) = 0.0470 M,
  • V = volume of acid = 16.0 mL,
  • C' = concentration of base (ammonia solution) and
  • V' = volume of base = 26.0 mL.
<h3>Concentration of ammonia solution</h3>

Making C' subject of the formula, we have

C' = CV/V'

Substituting the values of the variables into the equation, we have

C' = CV/V'

C' = 0.0470 M × 16.0 mL/26.0 mL

C' = 0.752 MmL/26.0 mL

C' = 0.0289 M

<h3>The concentration of acid at equivalence point</h3>

We know that the ion-product of water Kw is

Kw = [H⁺][OH⁻] =  where

  • [H⁺] = concentration of HCl at equivalence point,
  • [OH⁻] = C' = concentration of ammonia solution = 0.0289 M and
  • Kw = 1.01 × 10⁻¹⁴

Making [H⁺] subject of the formula, we have

[H⁺} = Kw/[OH⁻]

[H⁺] = 1.01 × 10⁻¹⁴/0.0289

[H⁺] = 34.95 × 10⁻¹⁴

[H⁺] = 3.495 × 10⁻¹³

<h3>pH at equivalence point</h3>

Since pH = -㏒[H⁺]

pH = -㏒[3.495 × 10⁻¹³]

pH = -㏒[3.495] + (-㏒10⁻¹³)

pH = -㏒[3.495] + [-13(-㏒10)]

pH = 13 - 0.5434

pH = 12.4566

pH ≅ 12.46

So, the pH at equivalence point is 12.46

Learn more about pH at equivalence point here:

brainly.com/question/25487920

3 0
2 years ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

3 0
3 years ago
Read 2 more answers
Which element is classified as a noble gas?
Fed [463]

Answer:

Noble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)

Explanation:

I found my answer on google

i hope it helps

please mark me as brainliest

6 0
3 years ago
How many moles are in 525 g of ammonia, NH3?<br> 8940.75 M<br> 0.03 M<br> 30.83 M
victus00 [196]

Answer:

30.83 M

Explanation:

17.03052 re in one mole. So, if you multiply it by 30.83, you will get 535 g of ammonia.

In fact, the detailed answer is 30.827009392549122.

3 0
2 years ago
Read 2 more answers
Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f
Marina86 [1]

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

     = -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15



4 0
3 years ago
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