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3 years ago

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

Answer: The volume of stock solution needed is 1.5 L

Explanation:

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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Define the following: Bronsted-Lowry acid - Lewis acid- Strong acid - (5 points) Problem 6: Consider the following acid base rea

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 $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}$