Number of carbon atom in 1 mol of carbon dioxide, CO2

For a particular first-order reaction, it takes 3.0 minutes for the concentration of the reactant to decrease to 25% of its init

grigory [225]

We will use this formula for first order:
㏑[A] = - Kt +Ao
when we have t (given)= 30 min = 30 x 60 = 1800 s (we here convert time from min to second.
then we assume that the initial concentration Ao = 1
and the concentration of A (final concentration = 0.25
So by substitution:
㏑(0.25) = - K * 1800 + ㏑(1)
1.39 = K * 1800
∴ K = 0.00077 s^-1 or 7.7 x 10^-4

5 0

4 years ago

2.70g of Zn (s) reacts with 50.0 mL of 1.00 M HCl solution to produce hydrogen gas according to the reaction. (R = 0.08206 L·atm

Anika [276]

Answer:

a. 0.0413 moles Zn

b. 0.0500 moles HCl

c. HCl is the limiting reactant

d. 0.0250 moles H₂

e. V = 0.56L

Explanation:

The reaction of Zn(s) with HCl is:

Zn(s) + 2HCl (aq) → ZnCl₂ (aq) + H₂(g)

Where 1 mole of Zn reacts with 2 moles of HCl.

a) To convert mass in grams to moles of a substance you need to use molar mass (Molar mass Zn: 65.38g/mol), thus:

2.70g Zn × (1mol / 65.38g) = 0.0413moles of Zn

b. Now, when you have a solution in molarity (Moles / L), you can know the moles of a volume of solution, thus:

Moles HCl:

50.0mL = 0.0500L × (1.00mol / L) = 0.0500 moles HCl

c. The limiting reactant is founded by using the chemical reaction as follows:

For a complete reaction of 0.0500 moles HCl you need:

0.0500 moles HCl × (1 mole Zn / 2 moles HCl) = 0.0250 moles Zn

As you have 0.0413 moles of Zn, and you need just 0.0250 moles for the complete reaction, Zn is the exces reactant and HCl is the limiting reactant

d.As HCl is limiting reactant and 2 moles of HCl react with 1 mole of H₂, moles of hydrogen formed are:

0.0500 moles HCl × (1 mole H₂ / 2 moles HCl) = 0.0250 moles H₂

e. Using PV = nRT, you can find volume of gas, thus:

PV = nRT

V = nRT / P

Where P is pressure 1atm at STP, n are moles, R is gas consant 0.08206Latm/molK and T is absolute temperature 273.15K at STP.

V = 0.0250molesₓ0.082atmL/molKₓ273.15K / 1atm

V = 0.56L

3 0

3 years ago

A gas is collected at 20.0 °C and 725.0 mm Hg. When the temperature is

krek1111 [17]

Answer:

676mmHg

Explanation:

Using the formula;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 725.0mmHg

P2 = ?

T1 = 20°C = 20 + 273 = 293K

T2 = 0°C = 0 + 273 = 273K

Using P1/T1 = P2/T2

725/293 = P2/273

Cross multiply

725 × 273 = 293 × P2

197925 = 293P2

P2 = 197925 ÷ 293

P2 = 676mmHg.

The resulting pressure is 676mmHg

3 0

3 years ago

Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is

mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol$