All categories

3 years ago

How many ml of 2.50 M NaOH solution are required to make a 525 mL of 0.150 M?

Chemistry

2 answers:

larisa [96]3 years ago

7 0

Answer:

31.5 mL of a 2.50M NaOH solution

Explanation:

Molarity (M) is an unit of concentration defined as moles of solute (In this case, NaOH), per liter of solvent. That is:

Molarity = moles solute / Liter solvent

If you want to make 525mL (0.525L) of a 0.150M of NaOH, you need:

0.525L × (0.150mol / L) = 0.07875 moles of NaOH

If you want to obtain these moles from a 2.50M NaOH solution:

0.07875mol NaOH × (1L / 2.50M) = 0.0315L = 31.5 mL of a 2.50M NaOH solution

stiks02 [169]3 years ago

3 0

Answer:

The correct answer is 31.5 ml

Explanation:

We have to dilute the more concentrated solution (2.50 M) to obtain a solution with a molarity of 0.150 M. We have the following data:

Initial concentration = Ci = 2.50 M

Final concentration = Cf = 0.150 M

Final volume = Vf = 525 ml

We consider the final and initial states and use the following expression to calculate the initial volume (Vi) in ml:

Ci x Vi = Cf x Vf

Vi= (Cf x Vf)/Ci = (0.150 M X 525 ml)/2.50 M = 31.5 ml

You might be interested in

Which energy changes are associated with a liquid boiling?

astraxan [27]

Answer:

Energy is absorbed, and potential energy increases

Explanation:

hope this helps

plz mark brainliest

5 0

2 years ago

What mass of potassium hypochlorite (FW-90.6 g/mol) must be added to 4.50 x 10 mL of water to give a solution with pH 10.20? [Ka

marta [7]

Answer : The mass of potassium hypochlorite is, 4.1 grams.

Explanation : Given,

pH = 10.20

Volume of water = $4.50\times 10^2ml=0.45L$

The decomposition of KClO will be :

$KClO\rightarrow K^++ClO^-$

Now the further reaction with water $(H_2O)$ to give,

$ClO^-+H_2O\rightarrow HClO+OH^-$

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$3.8=-\log [OH^-]$

$[OH^-]=1.58\times 10^{-4}M$

Now we have to calculate the base dissociation constant.

Formula used : $K_b=\frac{K_w}{K_a}$

Now put all the given values in this formula, we get :

$K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}$

Now we have to calculate the concentration of $ClO^-$ .

The equilibrium constant expression of the reaction is:

$K_b=\frac{[OH^-][HClO]}{[ClO^-]}$

As we know that, $[OH^-]=[HClO]=1.58\times 10^{-4}M$

$2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}$

$[ClO^-]=0.0999M$

Now we have to calculate the moles of $ClO^-$ .

$\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}$

$\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole$

As we know that, the number of moles of $ClO^-$ are equal to the number of moles of KClO.

So, the number of moles of KClO = 0.0449 mole

Now we have to calculate the mass of KClO.

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

$\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g$

Therefore, the mass of potassium hypochlorite is, 4.1 grams.

6 0

3 years ago

ASAP for brainlist first correct question

Ghella [55]

The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases

8 0

2 years ago

50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the

trapecia [35]

Answer:

-21 kJ·mol⁻¹

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50 50

c/mol·dm⁻³: 1.0 1.0

ΔT = 4.5 °C

C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}$

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

$\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0

0.050ΔH + 1883 + 225 = 0

0.050ΔH + 2108 = 0

0.050ΔH = -2108

ΔH = -2108/0.0500

= -42 000 J/mol

= -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0

3 years ago

When 0.25 moles of NH3 are produced according to the reaction below, how many moles of H2 must have reacted?

marissa [1.9K]

Answer:

Explanation:

by stoichiometry

nNH3/2=nH2/3

0.25/2=nH2/3

nH2=0.25×3/2

nH2=0.375 mol

so its 0.38 mol

6 0

3 years ago

Read 2 more answers