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Sergio039 [100]
3 years ago
5

how many mL of a 3.5 M sodium hydroxide will be needed to neutralize 15mL of a 4.3 M hydrochloric acid solution?

Chemistry
1 answer:
kobusy [5.1K]3 years ago
6 0
Step 1 : write a valanced equation..
NaOH + HCl 》NaCl + H2O

Step 2 : find the number of mole of HCl..
1000 ml ..contains 4.3 mole
15ml... (4.3÷1000)×15 =...

Stem 3 : use mole ratio....
HCl : NaOH
1 : 1
So mole is same as calculated above...

Step 4 :
3.5 mole of NaOH is in 1000ml
(4.3÷1000)×15 mole is in ....


Do the calculation
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The vapor pressure of methanol is 143 mmhg. identify the best reason to explain why methanol spontaneously evaporates in open ai
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Answer/Explanation:

Methanol has a molecular weight (32.04 g/mol), low-boiling point and because of its low boiling point, methanol readily evaporates at room temperature.

Under these specified non-standard conditions, the partial pressure of methanol is lower than its vapor pressure and this explains the reason for the spontaneous evaporation exhibited by methanol.

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Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2
Salsk061 [2.6K]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 51.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

2C(s)+2H_2(g)\rightarrow C_2H_4(g) \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.7kJ    ( × 2)

(2) H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_2=-285.9kJ     ( × 2)

(3) 2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)    \Delta H_3=-1411kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is 51.8 kJ.

6 0
3 years ago
If the a of a monoprotic weak acid is 2. 6×10−6, what is the ph of a 0. 33 m solution of this acid?
yawa3891 [41]

The pH of the monoprotic weak acid is 2.79.

<h3>What are weak acids?</h3>

The weak acids are the acids that do not fully dissociate into ions in the solution. Strong acids fully dissociate into ions.

The chemical reaction is HA(aq) ⇄ A⁻(aq) + H⁺(aq).

c (monoprotic acid) = 0.33 M.

Ka = 1.2·10⁻⁶

[A⁻] = [H⁺] = x

[HA] = 0.33 M - x

Ka = [A⁻]·[H⁺] / [HA]

2. 6 × 10⁻⁶ = x² / (0.33 M - x)

Solve quadratic equation: [H⁺] = 0.000524 M.

pH = -log[H⁺]

pH = -log(0.000524 M)

pH = 2.79

Thus,  the pH of the monoprotic weak acid is 2.79

To learn more about weak acids, refer to the below link:

brainly.com/question/13032224

#SPJ4

5 0
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