Which of the following is not an example of a base? KOH LiOH HCl NaOH

In order to prepare very dilute solutions, a lab technician chooses to perform a series of dilutions instead of measuring a very

SVETLANKA909090 [29]

Answer: The final concentration of potassium nitrate is $5.70\times 10^{-6}M$

Explanation:

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of potassium nitrate (solute) = 0.360 g

Molar mass of potassium nitrate = 101.1 g/mol

Volume of solution = 500.0 mL

Putting values in above equation, we get:

$\text{Molarity of }KNO_3=\frac{0.360\times 1000}{101.1\times 500.0}\\\\\text{Molarity of }KNO_3=7.12\times 10^{-3}M$

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

$M_1V_1=M_2V_2$ .......(1)

Calculating for first dilution:

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_1=7.12\times 10^{-3}M\\V_1=10mL\\M_2=?M\\V_2=500.0mL$

Putting values in equation 1, we get:

$7.12\times 10^{-3}\times 10=M_2\times 500\\\\M_2=\frac{7.12\times 10^{-3}\times 10}{500}=1.424\times 10^{-4}M$

Calculating for second dilution:

$M_2\text{ and }V_2$ are the molarity and volume of the concentrated $KNO_3$ solution

$M_3\text{ and }V_3$ are the molarity and volume of diluted $KNO_3$ solution

We are given:

$M_2=1.424\times 10^{-4}M\\V_2=10mL\\M_3=?M\\V_3=250.0mL$

Putting values in equation 1, we get:

$1.424\times 10^{-4}\times 10=M_3\times 250\\\\M_3=\frac{1.424\times 10^{-4}\times 10}{250}=5.70\times 10^{-6}M$

Hence, the final concentration of potassium nitrate is $5.70\times 10^{-6}M$

8 0

2 years ago

What is the first way in which biology researchers present the results of their latest research?

antoniya [11.8K]

Almost always, the first way in which biology researchers present the results of their latest research is to discuss the results of previous research that they want to build off of.

5 0

3 years ago

The last element in any period always has:

brilliants [131]

Answer: The correct option is The properties of a noble gas.

Explanation: There are 7 periods in the periodic table.

The last element of each period are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo).

The electronic configuration for Helium is $1s^2$ . For He, The outermost electrons are 2.

The electronic configuration for all the other elements is $ns^2ns^6$ ( where, n = 2, 3, 4, 5, 6 and 7 respectively). For all the other gases, the outermost electrons are 8.

All these elements have stable electronic configuration and are not reactive in nature. Hence, they are considered as noble gases.

Therefore, the last element of each period always have the properties of a noble gas.

4 0

3 years ago

Read 2 more answers

The first order reaction of methyl isonitrile to form acetonitrile has a rate constant of k = 1.7 × 10−15 at 298 K. What will ha

ludmilkaskok [199]

Refer to to this chart

Credits to https://image.slidesharecdn.com/nybf09-unit2slides26-57-100125181507-phpapp01/95/nyb-f09-unit-2-slid...

As temperature decreases, rate decreases.

As temp increases, k also increases.

3 0

3 years ago

Read 2 more answers

How many moles of CF4 can be produced when 8.95 mol C reacts with 7.88 mol F2?

Molodets [167]

The remaining moles of C is 5.01 moles while the remaining moles of F₂ is 0.

<h3>Reaction between Carbon and Fluorine </h3>

The reaction between carbon and Fluorine is given as;

C + 2F₂ -------> CF₄

1 : 2 1

from the reaction above,

2 moles of F₂ requires 1 mole of C

7.88 mole of F₂ will require: 7.88/2 = 3.94 moles of C and 3.94 moles of CF₄.

The remaining moles of C = 8.95 - 3.94 = 5.01 moles while the remaining moles of F₂ is 0.

Learn more about moles here: brainly.com/question/15356425

#SPJ1

8 0

2 years ago