2 What has been one effect of population growth on the physical environment?

Part B When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 22.8 g of carbon were burned in the pres

aksik [14]

<u>Answer:</u> The amount of carbon dioxide gas produced in the reaction is 83.6 grams

<u>Explanation:</u>

As, some amount of oxygen gas is left after the reaction is completed. So, it is present in excess and is considered as an excess reagent.

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

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 carbon = 22.8 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon}=\frac{22.8g}{12g/mol}=1.9mol$

The chemical equation for the reaction of carbon and oxygen gas follows:

$C+O_2\rightarrow CO_2$

By Stoichiometry of the reaction:

1 mole of carbon produces 1 mole of carbon dioxide gas

So, 1.9 moles of carbon will produce = $\frac{1}{1}\times 1.9=1.9moles$ of carbon dioxide gas

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 1.9 moles

Putting values in equation 1, we get:

$1.9mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(1.9mol\times 44g/mol)=83.6g$

Hence, the amount of carbon dioxide gas produced in the reaction is 83.6 grams

6 0

2 years ago

How do you calculate the volume of a cube? A. with a graduated cylindar in mL? B. with a triple beam balance in g? C. with a bea

Nuetrik [128]

Answer:

B

Explanation:

triple beam balance

sorry if i am wrong! can be brainliest?

5 0

2 years ago

Draw the major monobromination product formed by heating the following alkane with bromine.

yawa3891 [41]

A monobromination reaction of an alkane involves an alkane and bromine. The position of the hydrogen atom that will be substituted by the bromine free radical will depend on the order of the alkane. The bromine will attach to the carbon that has the most substituents.

8 0

3 years ago

H A and H B are both weak acids in water, and HA is a stronger acid than HB. Which of the following statements is correct? Selec

lubasha [3.4K]

Answer:

B is a stronger base than A^-, which is a stronger base than H2O, which is a stronger base than CI^-

Explanation:

The general equation for each acid is:

HA(aq) + H2O(ac) ⇄ H3O+(aq) + A-(aq)

HB(aq) + H2O(ac) ⇄ H3O+(aq) + B-(aq)

When these acids dissociate into its ions in water they lose a proton (H+), so they are proton donors (acids) and H2O is the proton acceptor (base). This reaction produces a conjugate acid and a conjugate base.

Conjugate base is what remains of the acid molecule after it loses a proton:

HA = acid A- Conjugate base

HB = acid B- Conjugate base

A conjugate acid is formed when the proton is transferred to the base

H2O = base H3O+ = Conjugate acid

The stronger acid will produce a weaker base. According to this, if HA is a stronger acid than HB, A- would be the weaker base (B- is the stronger base).

Compared with water, A- and B- are stronger bases because when they compete for a proton they have much greater affinity for H+ than water does and the equilibrium position will lie far to the left. (HA and HB are weak acids)

Finally Cl- is the weakest base because it comes after dissociation of HCl which is a strong acid

HCl(aq) + H2O → H3O+(aq) + Cl-(aq)

Note there is no double arrows, equilibrium lies far to the right. A strong acid yields a weak conjugate base it means one that has a low affinity for a proton.

4 0

3 years ago

A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t

jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:

protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺

protonated lidocaine lidocaine H₃O⁺

Initial(M) 0.22 0 0

Change -x +x +x

Equilibrium 0.22 - x x x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22

and solve for x. The pH will then be the negative log of this value.

8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0

3 years ago

2 What has been one effect of population growth on the physical environment? ​

2 What has been one effect of population growth on the physical environment?