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Alex787 [66]
2 years ago
9

Which of the following lists describes characteristics of a base?

Chemistry
2 answers:
soldi70 [24.7K]2 years ago
5 0

Answer: Bitter taste, high pH, and caustic

Explanation:

Acids are those substances which either donates hydrogen ions (H^+) when dissolved in water. They have pH ranging from 1 to 6.9.They are They are sour in taste. They dissolve metals to give hydrogen gas.Strong acids are caustic as they burn skin.

HX\rightarrow H^+X^-

Bases are those substances which either donates hydroxide ions (OH^-) when dissolved in water or donates a pair of electrons.  They have pH ranging from 7.1 to 14. They are bitter in taste. They are slippery in nature. Strong bases are caustic as they burn skin.

BOH\rightarrow B^++OH^-

suter [353]2 years ago
3 0

Answer:

bitter taste low ph and slipery

Explanation:

if its sour or dissolves metal it is an acid

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Methane (CH4) reacts with excess oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). What is the percent yield of c
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(29.8 g) / [0.184 mol (44.00964 g CO2/mol)] =0.832= 83.2% yield CO2

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A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6
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A student is given two 10g samples, each a mixture of only NaCl(s) and KCl(s) but in different proportions. Which of the followi
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A 10.0-ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the ph at the equivalence point? for
tatyana61 [14]
When the titration of HCN with NaOH is:

HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)

So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1

we need to get number of mmol of HCN = molarity * volume 

                      = 0.2 mmol / mL* 10 mL = 2 mmol

so the number of mmol of NaOH = 2 mmol according to the molar ratio

so, the volume of NaOH = moles/molarity

                                          = 2 mmol / 0.0998mL

                                          = 20 mL

and according to the molar ratio so, moles of CN- = 2 mmol

∴the molarity of CN- =  moles / total volume 

                                   = 2 mmol / (10mL + 20mL ) = 0.0662 M

when we have the value of PKa = 9.31 and we need to get Pkb

so, Pkb= 14 - Pka

            = 14 - 9.31 = 4.69 

when Pkb = -㏒Kb

         4.69 = -㏒ Kb 

∴ Kb = 2 x 10^-5

and when the dissociation reaction of CN- is:

CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq) 

by using the ICE table:

∴ the initials concentration are:

[CN-] = 0.0662 M

and [HCN] = [OH]- = 0 M

and the equilibrium concentrations are:

[CN-] = (0.0662- X)

[HCN] = [OH-]= X

when Kb expression = [HCN][OH-] /[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.0662 - X)

X = 0.00114 

∴[OH-] = X = 0.00114

when POH = -㏒[OH]

                    = -㏒ 0.00114

POH = 2.94

∴PH = 14 - 2.94 = 11.06



 

6 0
2 years ago
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