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KIM [24]
3 years ago
10

Net Ionic Equations for mixing Strong Acids with Strong Bases Consider a reaction between hydrochloric acid and potassium hydrox

ide.
a) In an aqueous solution of hydrochloric acid, check all of the major species found in solution (ignore the trace hydronium ions and hydroxide ions that would come from the autoionization of water). HCI (1) CIo (aq) CI (aq) OH (ag)H20 () H3o (aq)
b) In an aqueous solution of potassium hydroxide, check all of the major species found in solution (ignore the trace hydronium ons and hydroxide ions that would come from the autoionization of water). H20 () K+ (aq) OH (aq) Hyo (aq) KOH (5)
c) When the acid and base react together, they will neutralize each other to form water and a salt. Give the chemical formula for the salt formed. chemPad Help Greek acid and potassium hydroxide. Write it out in this order. Remember that, by convention, a net ionic equation has a single reaction arrow
d) When you cancel out the spectator ions, what is the net ionic equation that remains for the reaction between hydrochloric Help chemPad Oreek ▼
Chemistry
1 answer:
AleksandrR [38]3 years ago
4 0

Answer:

sorry but I am just answering the questions because I need points

Explanation:

thank you

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Which of the following would represent a single displacement reaction between Potassium Bromide (K^+Br^-) and Iodine (I), which
ra1l [238]

Answer:

Option : KBr + I -> KBr+I

Explanation:

Single-replacement reaction or single displacement reactions are a type of chemical reactions in which a whole compound reacts with an element in such a way that the element takes place of one of the compound's own elements and sets it free.

If we talk about KBr and I displacement reaction is not possible among these because Iodine is less reactive than Bromine that is why it will not react with KBr or replace Br.

                       KBr + I -> KBr+ I


Potassium Bromide + Iodine -> Potassium bromide + Iodine


Hope it help!

5 0
3 years ago
You have three drinks in front of you. you know how they are made, but are unsure which one has the highest concentration of jui
GuDViN [60]

(I think) you would have to divided the number i dont know the exact answer but ya hope this help
4 0
3 years ago
Read 2 more answers
ph ch3 h2nnh koh The Wolff-Kishner reaction involves the reaction of an aldehyde/ketone with hydrazine in the presence of KOH. T
Tatiana [17]

Hope this was helpful!

6 0
3 years ago
How many moles of sulfur dioxide (SO2) are required to produce 5.0 moles of sulfur (S) according to the following balanced equat
scoundrel [369]

Answer:

1.67 moles

Explanation:

From the balanced equation of reaction:

                     SO_2 + 2H_2S -> 3S + 2H_2O

1 mole of sulfur dioxide, SO2, is required to produce 3 moles of sulfur, S.

<em>If 1 mole SO2 = 3 moles S, then, how many moles of SO2 would be required for 5 moles S?</em>

     Moles of SO2 needed = 5 x 1/3

                   = 5/3 or 1.67 moles

Hence, <u>1.67 moles of SO2 would be required to produce 5.0 moles of S.</u>

3 0
2 years ago
Read 2 more answers
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
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