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Lilit [14]
2 years ago
14

Titration reveals that 11.6 mL of 3.0M sulfuric acid are required to neutralize the sodium hydroxide in 25.00mL of NaOH solution

. What is the molarity of the NaOH solution?
Chemistry
1 answer:
Deffense [45]2 years ago
6 0

Answer:

Explanation:

Molarity of acid(volume of acid)(# of H ions)= molarity of base(volume of base)(# of OH ions)

M(v)(#)=M(v)(#)

sulfuric acid    sodium hydroxide

H2SO4           NaOH

(3)(11.6)(2)=M(25)(1)

M=2.784

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Consider a saturated solution formed when 17.5 g of a solute dissolve in 28.3 g of a solvent, giving a total solution volume of
STALIN [3.7K]

Answer:

a) 38.2 % mass

b) 61.8 g solute/100 g solvent

c) 1.65 g/mL

Explanation:

Given the data:

mass of solute = 17.5 g

mass of solvent= 28.3 g

total solution volume= 27.8 mL

a)- mass percent= mass of solute/mass of solution x 100

mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g

mass % = 17.5 g/45.8 g x 100 = 38.2 % mass

b)- solubility = grams of solute/ 100 g solvent

                    = 17.5 g x (100 g /28.3 g solvent) = 61.8 g solute/100 g solvent  

c)- density = massof solution/total volumesolution  = 45.8 g/27.8 mL = 1.65 g/mL

7 0
2 years ago
Consider the solutions, 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3 and 0.04 m CaCl2. Which has (i) the highest osmotic pressure, (ii
Lana71 [14]

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

the number of particle produced by CaCl2 = 3

As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

5 0
2 years ago
he first-order rate constant for the gas-phase decomposition of dimethyl ether, (CH3)2O → CH4 + H2 + CO is 3.2 ✕ 10−4 s−1 at 450
seropon [69]

Answer:

0.290 atm is the pressure of the system after 7.7min

Explanation:

The general first-order rate constant is:

ln [A] = -kt + ln [A]₀

<em>Where [A] is concentration of A in time t,</em>

<em>K is rate constant, 3.2x10⁻⁴s⁻¹</em>

<em>[A]₀ is initial concentration = 0.336atm.</em>

<em />

7.7 min are:

7.7min * (60s / 1min) = 462s

Solving:

ln [A] = -kt + ln [A]₀

ln [A] = -<em>3.2x10⁻⁴s⁻¹*462s</em> + ln [0.336atm]

ln [A] = -1.238

[A] =

<h3>0.290 atm is the pressure of the system after 7.7min</h3>

<em />

6 0
2 years ago
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0
2 years ago
Read 2 more answers
A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution after the addition of 30.0 ml o
gizmo_the_mogwai [7]
From the molarity and volume of HClO4, we can determine how many moles of H+ we initially have: 
     0.18 M HClO4 * 0.100 L HClO4 = 0.018 moles H+

We can determine how many moles of OH- we have from the molarity and volume of LiOH:
     0.27 M LiOH * 0.030 L LiOH = 0.0081 moles OH-

When the HClO4 and LiOH neutralize each other, the remaining will be
     0.018 moles H+ - 0.0081 moles OH- = 0.0099 moles of excess H+

This means that the molarity [H+] will be
     [H+] = 0.0099 moles H+ / (0.100 L + 0.030 L) = 0.07615 M

The pH of the solution will therefore be
     pH = -log [H+] = -log 0.07615 = 1.12
6 0
2 years ago
Read 2 more answers
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