Balance this reaction: (_)NH3 + (_)H2SO4 = (_) (NH4)2SO4 1.2.2 2,1,1 2.2.1

Sort the phrases based on whether they describe or give an example of diffusion, facilitated diffusion, both, or neither. note:

Bingel [31]

Question:

a. Diffusion

b. Facilitated diffusion

c. Both

d. Neither

1. movement to area of lower concentration

2. movement across a membrane

3. steroid transport into cell

4. requires energy

5. movement assisted by proteins

6. glucose transport into cell

Answer:

The sorting is as follows

a. (1)

b. (5 and 6)

c. (1 and 2)

d. (4)

Explanation:

Diffusion is the movement of particles across a membrane from a high concentration region to one with a lower concentration of the diffusing substance

Here we have the correct sorting as follows

a. Diffusion

3. steroid transport into cell

b. Facilitated diffusion

5. movement assisted by proteins

6. glucose transport into cell

c. Both

1. movement to area of lower concentration

2. movement across a membrane

d. Neither

4. requires energy

8 0

3 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

slega [8]

Answer:

The catalyzed reaction will take time of $5.11\times 10^{-8} years$ .

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate of reaction with catalyst

$K_1$ = rate of reaction without catalyst

$Ea_2$ = activation energy with catalyst = 59.0 kJ/mol = 59000 J/mol

$Ea_1$ = activation energy without catalyst = 184 kJ/mol = 184000 J/mol

R = gas constant

T = temperature = $600K$

Now put all the given values in this formula, we get:

$\frac{K_2}{K_1}=e^{\frac{184,000 kJ-59000 kJ}{R\times 300}}=7.632\times 10^{10}$

The reaction enhances by $7.632\times 10^{10}$ when catalyst is present.

Time taken by reaction without catalyzed = 3900 years

Time taken by reaction with catalyzed = x

$x=\frac{3900 year}{7.632\times 10^{10}}=5.11\times 10^{-8} years$

The catalyzed reaction will take time of $5.11\times 10^{-8} years$ .

8 0

3 years ago

How many TOTAL carbon atoms are their in this equation? A. 1 B. 2 B.4 B. 6

telo118 [61]

What equation, love?

8 0

4 years ago

Read 2 more answers

Take a close look at this diagram on the slide which form of government is represented by this diagram

Allushta [10]

Which diagram? There isn’t any there lol

8 0

3 years ago

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH = –1.37 × 103 kJ For the combustion of ethyl alcohol as described in the above equati

babymother [125]

Answer:

The true statements are: I. The reaction is exothermic.

II. The enthalpy change would be different if gaseous water was produced.

Explanation:

The given chemical reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH= -1.37×10³kJ

1. In an exothermic reaction, heat or energy is released from the system to the surrounding. Thus for an exothermic process the change in enthalpy is less than 0 or negative (ΔH < 0) .

Since the enthalpy change for a combustion reaction is negative. Therefore, the given reaction is exothermic.

2. The change in enthalpy (ΔH) of a reaction is equal to difference of the sum of standard enthalpy of formation (ΔHf°) of the products and the reactants.

ΔHr° = ∑ n.ΔHf°(products) − ∑ n.ΔHf°(reactants)

As the value of ΔHf° of water in gaseous state and liquid state is not the same.

Therefore, the enthalpy change of the reaction will be different, if gaseous water was present instead of liquid water.

3. An oxidation-reduction reaction or a redox reaction involves simultaneous reduction and oxidation processes.

The given chemical reaction, represents the combustion reaction of ethanol.

Since combustion reactions are redox reactions. Therefore, the given combustion reaction is an oxidation-reduction reaction.

4. According to the ideal gas equation: P.V =n.R.T

Volume (V) ∝ n (number of moles of gas)

Since the number of moles (n) of gaseous reactants is 3 and number of moles of gaseous (n) products is 2.

Therefore, the volume occupied by 3 moles of the reactant gaseous molecules will be more than 2 moles product gaseous molecules.

3 0

3 years ago

Balance this reaction: (_)NH3 + (_)H2SO4 = (_) (NH4)2SO4 1.2.2 2,1,1 2.2.1​

Balance this reaction: (_)NH3 + (_)H2SO4 = (_) (NH4)2SO4 1.2.2 2,1,1 2.2.1