Whatisthechemicalequationforthereactionofaqueoussodiumhydroxidewithaqueoushydrochloricacid<br>

1) Sodium hydroxide is deliquescent. A sample of 3.0 g was dissolved in 100 mL; 10 mL was titrated with 34.9 mL HCl 0.2 M. What

ELEN [110]

From the calculation, the percentage of water in the sodium hydroxide sample is 7%.

<h3>What is neutralization?</h3>

The term neutralization has to do with the reaction between an acid and a base to yiled salt and water.

Now we have to apply the titration formula;

CAVA/CBVB = NA/NB

CA = concentration of acid

CB = concentration of base

VA = volume of acid

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

The reaction equation is; HCl + NaOH ----->NaCl + H2O

CAVANB = CBVBNA

CB = CAVANB /VBNA

CB = 34.9 * 0.2 M * 1/10 * 1

CB = 0.698 M

Number of moles = Conncentration * volume

= 0.698 M * 100/1000 L = 0.0698 moles

Mass = Number of moles * molar mass

Mass = 0.0698 moles * 40 g/mol = 2.79 g

percent of NaOH = 2.79 g/ 3g * 100/1 = 93%

Percent of water = 100- 93 = 7%

Learn more about neutralization: brainly.com/question/15395418

5 0

2 years ago

Molecular iodine, I2(g), dissociates into iodine atoms at 625 K with a first-order rate constant of 0.271 s−1 .

katrin [286]

The correct is this because they said

4 0

2 years ago

Be sure to answer all parts. Determine the electron-group arrangement, molecular shape, and ideal bond angle for the following m

choli [55]

Answer:

trigonal planar

Explanation:

The molecule SO3 is of the type AX3. The molecule is symmetrical and non polar.

There are three regions of electron density in the molecule. This corresponds to a trigonal planar geometry. This means that the three oxygen atoms are arranged at the corners of a triangle. The bond angle is 120 degrees.

7 0

3 years ago

Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0

olya-2409 [2.1K]

Answer:

$pH=-1.37$

Explanation:

We are given that 25 mL of 0.10 M $CH_3COOH$ is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of $CH_3COOH=25mL=0.025 L$

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL= $\frac{35}{1000}=0.035 L$

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of $CH_3COOH$ =0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+= $0.10\times 0.025$ =0.0025 moles

Number of moles of $OH^-=0.10\times 0.01$ =0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+= $\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L$

pH=-log [H+]=-log [4.28 $\times 10^{-2}$ ]=-log4.28+2 log 10=-0.631+2

$pH=-1.37$

6 0

3 years ago

Indicar la cantidad de sustancia en: a) 2.0L de un gas en C.N; b) 22.4 mL de un gas a 755 torr y 26 grados celsius?

nekit [7.7K]

Answer:

El termopar B presenta un mayor grado de dispersión y también es más preciso. ... (c) La estimación para T = 175 ° C es probablemente la más cercana al valor real, porque el ... (cm3). Flujo de masa. Velocidad. (kg / min). Diferencia. Duplicar. (Di). Yo y yo. 2. 1 ... atm de gas. 2. 2. 2 f. 3. 2 f f. 30 14,7 lb 20 pulg. 4 14,7 lb 24 pulg 392 lb 7,00 10 lb pulg.

8 0

2 years ago

Whatisthechemicalequationforthereactionofaqueoussodiumhydroxidewithaqueoushydrochloricacid​

Whatisthechemicalequationforthereactionofaqueoussodiumhydroxidewithaqueoushydrochloricacid