What are centimeters,kilometers,and millimeters

How many molecules are in 23 Moles of oxygen o2

loris [4]

Answer:

1.38×10^25 molecules

Explanation:

Applying n= (no. of molcules)/NA

23 = N/6.02×10^23

= 1.38×10^25 molecules

6 0

3 years ago

If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T

Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

<u>Before adding HCl,</u>

7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔ log([Conjugate base]/[Acid]) = 0

⇔ [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

<u>After adding HCl (3.00 mL, 1.00 M)</u>

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

= 7.55 + log(0.003 / 0.009) = 7.07

8 0

3 years ago

Diethyl ether is produced from ethanol according to the following equation:

juin [17]

The balanced chemical equation is given as:

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.

Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3

% yield = .775 = actual yield / 264.66

actual yield = 205.11 g CH3CH2OCH2CH3

6 0

3 years ago

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

stellarik [79]

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^{2}$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^{2}$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

4 0

4 years ago

Draw the Lewis dot structure for RbIO2. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atom

Rzqust [24]

Answer : The Lewis-dot structure of $RbIO_2$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $RbIO_2$

As we know that rubidium has '1' valence electrons, iodine has '7' valence electrons and oxygen has '6' valence electrons.

Therefore, the total number of valence electrons in $RbIO_2$ = 1 + 7 + 2(6) = 20

As we know that $RbIO_2$ is an ionic compound because it is formed by the transfer of electron takes place from metal to non-metal element.

7 0

3 years ago