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Rama09 [41]
2 years ago
15

What is the minimum volume of 5.50 M HCl necessary to neutralize completely the hydroxide in 718.0 mL of 0.183 M NaOH

Chemistry
1 answer:
Deffense [45]2 years ago
6 0

The volume of HCl required is 23.89 mL

Calculation of volume:

The reaction:

HCl + NaOH \rightarrow NaCl + H_2O

As HCl and NaOH react in 1 : 1 ratio.

Volume of NaOH= 718 mL

Concentration= 0.183M

Volume of HCl= ?

Concentration= 5.50M

Using the dilution formula:

M_1\times V_1(NaOH)= M_2\times V_2(HCl)

0.183\times 718= 5.50 \times V_2\\V_2=\frac{131.394}{5.50} \\V_2 = 23.89 \,mL

Therefore,

Volume of HCl required will be 23.89 mL.

Learn more about neutralization reaction here,

brainly.com/question/1822651

#SPJ4

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Mechanical twinning occurs in metals having which type(s) of crystal structure(s)?
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according to the balanced chemical equation below,how many grams of h20 are produced if 3.98g of co2 were produced. 2C18H18 + 25
telo118 [61]

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Primero debes usar los gramos de co2 y luego buscar su peso molecular, luego de eso usar la relación de moles entre CO2 y H2O y por último buscar el pm del H2O pata ver cuantos gramos de produce.

Explanation:

3 0
3 years ago
A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75
Stolb23 [73]

Answer:

5 moles of Argon is present in the mixture.

Explanation:

Total pressure of the gaseous mixture = 4 atm

Total number of moles = 16

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Hence:

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