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VLD [36.1K]
3 years ago
8

What volume of a 3.0 M stock solution of H2SO4 is needed to prepare 2.8 L of a 1.6 M H2SO4 solution?

Chemistry
2 answers:
DiKsa [7]3 years ago
5 0

<u>Answer:</u> The volume of stock solution needed is 1.5 L

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the stock solution

M_2\text{ and }V_2 are the molarity and volume of diluted solution

We are given:

M_1=3M\\V_1=?L\\M_2=1.6M\\V_2=2.8L

Putting values in above equation, we get:

3\times V_1=1.6\times 2.8\\\\V_1=1.5L

Hence, the volume of stock solution needed is 1.5 L

Georgia [21]3 years ago
4 0
Use M1V1 = M2V2 to solve
3(V1) = 2.8 * 1.6
3(V1) = 4.48
V1 = 1.493 L of stock solution
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Allushta [10]

Answer: Yes, HCl is a strong acid.

acid = HCl , conjugate base = Cl^- , base = H_2O, conjugate acid = H_3O^+

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Yes HCl is a strong acid as it completely dissociates in water to give H^+ ions.

HCl\rightarrow H^++Cl^-

For the given chemical equation:

HCl+H_2O\rightarrow H_3O^-+Cl^-

Here, HCl is loosing a proton, thus it is considered as an acid and after losing a proton, it forms Cl^- which is a conjugate base.

And, H_2O is gaining a proton, thus it is considered as a base and after gaining a proton, it forms H_3O^+ which is a conjugate acid.

Thus acid =  HCl

conjugate base = Cl^-

base = H_2O

conjugate acid = H_3O^+.

8 0
3 years ago
At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

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3 years ago
In which pair is the one on the left larger than the one on the right?
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3 years ago
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3 0
2 years ago
If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of
Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.1\ g}{186.04\ g/mol}

Moles\ of\ ferrocene= 0.0005375\ mol

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

\rho=\frac{Mass}{Volume}

or,  

Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.044\ g}{78.49\ g/mol}

Moles\ of\ acetyl\ chloride= 0.0005606\ mol

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms  0.0005375 mole of monoacetylferrocene

Moles of product formed =  0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0
3 years ago
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