Why do we use anhydrous diethyl ether? Choose the right answer.

Jim, Jill, Robert, and Kim each run paper chromatography on an unknown aqueous mixture. Jim gets a red band and a blue band, Jil

Romashka-Z-Leto [24]

Answer:

Robert

Explanation:

There is not more than one colour

4 0

2 years ago

Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 oC) in the absence of a ca

Oksi-84 [34.3K]

Answer:

30 kJ

Explanation:

Arrhenius equation is given by:

$k=Aexp(-Ea/RT)\\$

Here, k is rate constant, A is Pre-exponential factor, Ea is activation energy and T is temperature.

taking natural log of both side

ln k = ln A - Ea/RT

In Arrhenius equation, A, R and T are constant.

Therefore,

$ln\frac{k_2}{k_1} =\frac{Ea_1-Ea_2}{RT}$

$Ea_1-Ea_2$ is the lowering in activation energy by enzyme,

R = 8.314 J/mol.K

T = 37°C + 273.15 = 310 K

$\frac{k_2}{k_1} =1\times 10^5$

$ln 1\times 10^5 =\frac{Ea_1-Ea_2}{RT}\\{Ea_1-Ea_2} = 11.512 \times 8.314 \times 310\\=29670\ J\\=30\ kJ$

4 0

3 years ago

What mass, in grams, of CO2 and H2O is formed from 2.55 mol of propane?

oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Being the molar mass of each compound:

C₃H₈: 44 g/mole
O₂: 16 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

C₃H₈: 1 mole* 44 g/mole= 44 grams
O₂: 5 moles* 16 g/mole= 80 grams
CO₂: 3 moles* 44 g/mole= 132 grams
H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

$mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}$

mass of CO₂= 336.6 grams

If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

$mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}$

mass of H₂O= 183.6 grams

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

3 0

3 years ago

I need help please please

inna [77]

Answer:

B should be the answer, and ur low-key valid lol

Explanation:

6 0

3 years ago

A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what

Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. $[a]_{t}$

Calculations:

The second order rate equation is:

$1/[a]_{t} = kt +1/[a]$

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M

7 0

3 years ago