When Acid is added to the buffer, base within the buffer reacts with added

What is the molarity of a 250.0 milliliter aqueous solution of sodium hydroxide that contains 15.5 grams of solute? 6.2 M

Alchen [17]

Moles = 15.5 g / 40 g/mol = 0.3875 mol

M = 0.3875 mol / 0.250 L = 1.55M

8 0

3 years ago

Read 2 more answers

Frayer model for matter

givi [52]

A model showing that gases are made from the matter of particles that are too small to see and are moving freely around in space can explain many observations.

6 0

2 years ago

PLEASE HELPPPP!!!!!

Likurg_2 [28]

Answer : The mass of nitric acid is, 214.234 grams.

Solution : Given,

Moles of nitric acid = 3.4 moles

Molar mass of nitric acid = 63.01 g/mole

Formula used :

$\text{Mass of }HNO_3=\text{Moles of }HNO_3\times \text{Molar mass of }HNO_3$

Now put all the given values in this formula, we get the mass of nitric acid.

$\text{Mass of }HNO_3=(3.4moles)\times (63..01g/mole)=214.234g$

Therefore, the mass of nitric acid is, 214.234 grams.

6 0

3 years ago

Determine a massa, em kg, de um material que está contida em um volume de 18L. Sabe-se que a densidade do material é de 0,9 g/cm

suter [353]

The question in English is "Determine the mass, in kg, of a material that is contained in a volume of 18L. It is known that the material density is 0.9 g/cm 3"

Answer:

We can use a simple equation to solve this problem. 
 d = m/v

Where d is the density, m is the mass and v is the volume.

d = 0.9 g/cm³
m = ?
v = 18 L = 18 x 10³ cm³

By applying the equation,
 0.9 g/cm³ = m / 18 x 10³ cm³
m = 0.9 g/cm³ x 18 x 10³ cm³
 m = 16200 g
m = 16.2 kg

Hence, the mass of 18 L of material is 16.2 kg.

6 0

2 years ago

A 1.00 L buffer solution is 0.112 M in acetic acid and 0.112 M in sodium acetate. Acetic acid has a pKa of 4.74. What is the pH

topjm [15]

Answer:

ΔpH = 1.25

Explanation:

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.112] / [0.112]

pH = 4.74

The reaction of sodium acetate (CH₃COONa) with HCl is:

CH₃COONa + HCl → CH₃COOH + NaCl

Producing acetic acid, CH₃COOH.

If 0.1mol of HCl reacts the final moles of CH₃COONa are:

0.112mol - 0,1 mol = 0.012mol

Moles of acetic acid are:

0.112mol + 0,1 mol = 0.212mol

Using Henderson-Hasselbalch formula:

pH = pka + log [CH₃COONa] / [CH₃COOH]

pH = 4.74 + log [0.012] / [0.212]

pH = 3.49

Change in pH, ΔpH = 4.74 - 3.49 = 1.25

I hope it helps!

6 0

2 years ago