Provide a title short explanation or description accompanying an illustration or a photograph

A chemist dilutes a 1.0 ml sample of 2.0 m kno3 by adding water to it. if the concentration of the solution that is obtained is

Nitella [24]

A chemist dilutes a 1.0 ml sample of 2.0M KNO₃ by adding water to it. If the concentration of the solution that is obtained is 0.0080 m, then its volume is 250mL.

<h3>How do we calculate volume?</h3>

Volume for the given equation will be calculated by using the below equation as:

M₁V₁ = M₂V₂, where

M₁ = molarity of KNO₃ = 2M

V₁ = volume of KNO₃ = 1mL

M₂ = molarity of final solution = 0.0080M

V₂ = volume of final solution = ?

On putting these values on the above equation, we get

V₂ = (2)(1) / (0.0080) = 250mL

Hence option (2) is correct.

To know more about molarity, visit the below link:

brainly.com/question/24305514

8 0

2 years ago

natural rubidium has the average mass of 85.4678 and is composed of isotopes 85 Rb(mass = 84.9117) and 87 Rb. The ratio of atoms

mote1985 [20]

Answer:

Mass of Rb-87 is 86.913 amu.

Explanation:

Given data:

Average mass of rubidium = 85.4678 amu

Mass of Rb-85 = 84.9117

Ratio of 85Rb/87Rb in natural rubidium = 2.591

Mass of Rb = ?

Solution:

The ration of both isotope is 2.591 to 1. Which means that for 2.591 atoms of Rb-85 there is one Rb-87.

For 100% naturally occurring Rb = 2.591 + 1 = 3.591

% abundance of Rb-85 = 2.591/ 3.591 = 0.722

% abundance of Rb-87 = 1 - 0.722= 0.278

84.9117 × 0.722 + X × 0.278 = 85.4678

61.306 + X × 0.278 = 85.4678

X × 0.278 = 85.4678 - 61.306

X × 0.278 = 24.1618

X = 24.1618 / 0.278

X = 86.913 amu

8 0

3 years ago

An electric current can be measured with a device called?

worty [1.4K]

Answer: Ammeter

Explanation:

7 0

2 years ago

Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating

kvv77 [185]

<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

$1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol$

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

$\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }$

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = $\frac{1}{1}\times 0.0295=0.0295moles$ of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g$

To calculate the percentage yield of iodo-salicylamide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

$\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%$

Hence, the percent yield of the reaction is 68.68%.

4 0

3 years ago

Question 11 please HELP

Lena [83]

The answer is A or B, I would put B.

6 0

3 years ago

Provide a title short explanation or description accompanying an illustration or a photograph ​

Provide a title short explanation or description accompanying an illustration or a photograph