Why does naphthol react with sodium hydroxide?

Calculate the number of hydrogen atoms present in 40g of urea, (NH2)2CO

tekilochka [14]

Answer: There are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

Explanation:

Given: Mass of urea = 40 g

Number of moles is the mass of substance divided by its molar mass.

First, moles of urea (molar mass = 60 g/mol) are calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{40 g}{60 g/mol}\\= 0.67 mol$

According to the mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

So, the number of atoms present in 0.67 moles are as follows.

$0.67 mol \times 6.022 \times 10^{23} atoms/mol\\= 4.035 \times 10^{23} atoms$

In a molecule of urea there are 4 hydrogen atoms. Hence, number of hydrogen atoms present in 40 g of urea is as follows.

$4 \times 4.035 \times 10^{23} atoms\\= 16.14 \times 10^{23} atoms$

Thus, we can conclude that there are $16.14 \times 10^{23}$ atoms of hydrogen are present in 40g of urea, $(NH_{2})_{2}CO$ .

7 0

2 years ago

A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add

erma4kov [3.2K]

Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

47.2 mL * 150 mM = 24.0 mM * V₂

V₂ = 295 mL

And <u>converting into L </u>becomes:

295 mL * $\frac{1 L}{1000mL}$ = 0.295 L

6 0

3 years ago

ou will prepare 250-mL of this solution using a 30% (m/v) NaOH stock solution. How many mL of the NaOH stock solution will you n

ArbitrLikvidat [17]

Answer:

$\boxed{\text{3.3 mL}}$

Explanation:

You must convert 30 % (m/v) to a molar concentration.

Assume 1 L of solution.

1. Mass of NaOH

$\text{Mass of NaOH} = \text{1000 mL solution } \times \dfrac{\text{30 g NaOH}}{\text{100 mL solution}} = \text{300 g NaOH}$

2. Moles of NaOH

$\text{Moles of NaOH} = \text{300 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40.00 g NaOH}} = \text{7.50 mol NaOH}$

3. Molar concentration of NaOH

$c= \dfrac{\text{moles}}{\text{litres}} = \dfrac{\text{7.50 mol}}{\text{1 L}} = \text{7.50 mol/L}$

4. Volume of NaOH

Now that you know the concentration, you can use the dilution formula .

$c_{1}V_{1} = c_{2}V_{2}$

to calculate the volume of stock solution.

Data:

c₁ = 7.50 mol·L⁻¹; V₁ = ?

c₂ = 0.1 mol·L⁻¹; V₂ = 250 mL

Calculations:

(a) Convert millilitres to litres

$V = \text{250 mL} \times \dfrac{ \text{1 L}}{\text{1000 mL}} = \text{0.250 L}$

(b) Calculate the volume of dilute solution

$\begin{array}{rcl}7.50V_{1} & = & 0.1 \times 0.250\\7.50V_{1} &= & 0.0250\\V_{1} & = & \text{0.0033 L}\\& = & \textbf{3.3 mL}\\\end{array}$

$\text{You will need $\boxed{\textbf{3.3 mL}}$ of the stock solution.}$

4 0

2 years ago

ALSO NEED THIS ANSWERED PRETTY SOON I THINK IK WHAT IT IS BUT NOT SURE HELP!

Illusion [34]

Answer:

I think it's B but I could be wrong so really sorry if I am

3 0

1 year ago

Read 2 more answers

Charles law describes the relationship of the volume and temperature of gas at a constant mass and pressure. According to this l

zavuch27 [327]

Answer: Charles's law states that the volume of a given amount of gas is directly proportional to its temperature on the kelvin scale when the pressure is held constant. with k being a proportionality constant that depends on the amount and pressure of the gas.

Explanation:

6 0

2 years ago