Facilitated diffusion moves large molecules through _______

The characteristic odor of pineapple is due to ethyl butyrate, an organic compound which contains only carbon, hydrogen and oxyg

Trava [24]

Answer:

C3H6O

Explanation:

Data obtained from the question. This includes:

Carbon (C) = 0.62069 g

Hydrogen (H) = 0.103448 g

Oxygen (O) = 0.275862 g

The empirical formula can be obtained as follow:

Step 1:

Divide by their molar mass.

C = 0.62069 / 12 = 0.0517

H = 0.103448 / 1 = 0.103448

O = 0.275862 / 16 = 0.0172

Step 2:

Divide by the smallest number.

C = 0.0517 / 0.0172 = 3

H = 0.103448 / 0.0172 = 6

O = 0.0172 / 0.0172 = 1

Step 3:

Writing the empirical formula.

The empirical formula is C3H6O

3 0

3 years ago

Read 2 more answers

If the heating curve is reversed, what describes the boiling point?

GalinKa [24]

Boiling is the process of converting a substance from liquid state to gaseous state. If the heating curve is reversed, the process also is reversed from converting gaseous state to liquid state. In this case, the reverse of boiling is condensation. So the answer is point of condensation.

5 0

3 years ago

A 44.0 ml volume of 0.25 m hbr is titrated with 0.50 m koh. calculate the ph after addition of 22.0 ml of koh at 25 ∘c.

Helga [31]

The reaction between KOH and HBr is as follows ;
KOH + HBr ---> H₂O + KBr
Stoichiometry of base to acid is 1:1 molar ratio
Both are strong acid and strong base therefore complete ionization takes place
The number of KOH moles added - 0.50 M / 1000 mL/L x 22 mL = 0.011 mol
the number of HBr moles - 0.25 M /1000 mL/L x 44 mL = 0.011 mol
the number of H⁺ ions and OH⁻ ions are equal therefore the whole amount of acid has been completely neutralised by base.
No remaining acid nor base, therefore solution is neutral.
pH = 7
thats the pH value for a neutral solution

4 0

4 years ago

It was found that 2.35 g of a compound of phosphorus and chlorine contained 0.539 g of phosphorus. What are the percentages by m

igor_vitrenko [27]

Answer:

Explanation:

P = 2.35g

Cl= 0.539 g

% MASS = mass of X/ mass of a compound

% mass of P = 2.35 / (2.35+ 0.539) = 81.34%

% mass of Cl = 0.539 /(2.35+ 0.539) = 18.66 %

3 0

4 years ago

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temper

LUCKY_DIMON [66]

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

$t_{1/2}=\frac{1}{k\times a_0}$

k = rate constant =?

$a_0$ = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

$\frac{1}{a}=kt+\frac{1}{a_0}$

a) Initial concentration of XY = $a_o=0.100 M$

Rate constant of the reaction = k = $6.96\times 10^{-3} M^{-1} s^{-1}$

Half life of the reaction is:

$t_{1/2}=\frac{1}{6.96\times 10^{-3} M^{-1} s^{-1}\times 0.100 M}$

$=1,436.78 s$

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

$\frac{1}{0.0125 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.100M}$

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

$\frac{1}{0.025 M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t = $6.20\times 10^{-2} M$

$\frac{1}{6.20\times 10^{-2} M}=6.96\times 10^{-3} M^{-1} s^{-1}\times t+\frac{1}{0.200M}$

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become $6.20\times 10^{-2} M$ .

e) Initial concentration of $SO_2Cl_2= 0.050 M$

Final concentration after time t = x

t = 55.0 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 55.0 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

$\frac{1}{x}=6.96\times 10^{-3} M^{-1} s^{-1}\times 500 s+\frac{1}{0.050 M}$

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.

7 0

3 years ago

Facilitated diffusion moves large molecules through ________.