There are different atoms and each type of atom is called an

Black_prince [1.1K]

4 NH₃ + 3O₂ --> 2N₂ + 6H₂O

First, make sure that this is a balanced equation.
There are 4 moles of nitrogen on the left side, and 4 moles of nitrogen on the right side.
There are 12 moles of hydrogen on the left side, and 12 moles of hydrogen on the right side.
There are 6 moles of oxygen on the left side, and 6 moles of oxygen on the right side.

The equation is therefore balanced, and we may proceed.

a) the mole ratio for NH₃ to N₂ is 4 to 2, which can be simplified to 2:1 or 2/1.

b) the mole ratio for H₂O to O₂ is 6 to 3, which can be simplified to 2:1 or 2/1.

6 0

3 years ago

Now select a longer period, say the fourth period, starting with the element in the 1A group. Again, go from left to right. What

Inessa [10]

Answer:

The pattern of change in the electron configuration increases from left to right across the period.

Explanation:

In the periodic table, elements having the same number of electrons in the outermost shell of their atoms are placed over one another in vertical columns. Each of the vertical columns is known as a group or family. while each of the resulting horizontal rows is known as a period or row.

There are 18 vertical columns and seven periods in the periodic table. Each period begins with an atom having a valence electron and ends with an atom having a complete outer shell structure of an inert gas.

The fourth period of the periodic table (d-block) consists of the transition elements.

The electron distribution into the energy levels of sublevels of atoms can be shown in any of the tree important electronic configuration.

Electronic configuration shows the sequence of filling electrons into the orbitals of the sublevels as guided by some principles.

The maximum number of electrons in the orbitals of sublevels are two for s-sublevel(one orbital) ; six for p-sublevel( three orbital); ten for d-sublevel ( five orbitals) and fourteen for f-sublevel( seven orbitals). This indicates that the maximum number of electrons in an orbital is two

In the Periodic table ,The sublevels with lower energies are filled up before those with higher energies. One important thing about this principle is that the sublevels do not fill up in numerical order. The pattern of filling is as follows:

1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f etc.

5 0

3 years ago

Calculate the volume of compound Y needed to make a 350 pg/ml treatment solution in 2 ml of 1x PBS using a stock solution contai

harina [27]

Answer:

14 mL

Explanation:

To prepare a solution by a concentrated solution, we must use the equation:

C1xV1 = C2xV2, where C is the concentration, V is the volume, 1 is the initial solution and 2 the final solution.

The final solution must have 2 mL and a concentration of 350 pg/mL, and the initial solution has a concentration of 50 pg/mL.

Then:

50xV1 = 350x2

50xV1 = 700

V1 = 700/50

V1 = 14 mL

4 0

3 years ago

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole r

spin [16.1K]

Answer: The theoretical yield of acetanilide is 6.5 grams.

Explanation:

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

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

For aniline:

Given mass of aniline = $4.50\times 10^0=4.50g$ (We know that: $10^0=1$ )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

$\text{Moles of aniline}=\frac{4.50g}{93.13g/mol}=0.048mol$

For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

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

Volume of acetic anhydride = $(1.25\times \text{Mass of aniline})=1.25\times 4.50=5.625mL$

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

$1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL\times 5.625mL)=6.08g$

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of acetic anhydride}=\frac{6.08g}{102.1g/mol}=0.06mol$

The chemical equation for the reaction of aniline and acetic anhydride follows:

$C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH$

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with = $\frac{1}{1}\times 0.048=0.048mol$ of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce = $\frac{1}{1}\times 0.048=0.048mol$ of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

$0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol\times 135.17g/mol)=6.5g$

Hence, the theoretical yield of acetanilide is 6.5 grams.

3 0

3 years ago

A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.

yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH since after reaction we will have both the conjugate base lactate anion and unreacted weak lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA = (20.0 mL x 1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴ ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.0005 0

after rxn 0.002-0.0005 0 0.0005

0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.001 0

after rxn 0.002-0.001 0 0.001

0.001

pH = 3.85 + log(0.001/0.001) = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.002 0

after rxn 0 0 0.002

We are at the neutralization point and we do not have a buffer anymore, instead we just have a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻] = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log ( 1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

HA + NaOH ⇒ A⁻ + H₂O

before rxn 0.002 0.0025 0

after rxn 0 0.0005 0.0005

Now here what we have is the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0

3 years ago

There are different atoms and each type of atom is called an​

There are different atoms and each type of atom is called an