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Alex Ar [27]
2 years ago
15

HELP I NEED HELP ASAPP

Chemistry
1 answer:
Effectus [21]2 years ago
4 0

Answer:

B = A/DH – C

Explanation:

From the question given above, we obtained:

A = D • H(B + C)

Thus, we can obtain B in terms of D, H, A and C by doing the following:

A = D • H(B + C)

A = DH(B + C)

Divide both side by DH

A/DH = B + C

Subtract C from both side

A/DH – C = B + C – C

A/DH – C = B

B = A/DH – C

You might be interested in
Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0
3 years ago
Read 2 more answers
A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop in 7.5 seconds? What is the cars acc
Marianna [84]

Answer:

The acceleration is: 4m/s^2

Explanation:

Given

u = 30.0m/s --- The initial velocity

t = 7.5s --- time

v = 0m/s -- The final velocity

Required

Determine the acceleration

To do this, we make use of the first equation of motion

v = u - at

We used negative because the car was coming to stop.

This gives:

0 = 30 - 7.5 * a

0 = 30 - 7.5a

Collect like terms

7.5a = 30

Solve for a

a = \frac{30}{7.5}

a = 4m/s^2

6 0
3 years ago
ASAP PLLEEASE HURYYYYYYYYYYRRRRRRRRRRRRYY (NO FILES FOR ME NEEDING TO DOWNLAOD PLEASE!)
pickupchik [31]

Answer:

1.Fill in the blank with the correct term to make the statement true.

During the (night)

at the beach, the land cools down

(slower)

than the

(ocean).

Because of this, the air over the land is

(Warmer) than the air over the ocean. The

(More)

dense air rises over the ocean and the cooler,

(More)

dense air blows in from over the land creating a

(sea)

breeze.

2.Which statement is true about winds?

Winds always blow from low to high pressure.

true

Winds always blow from water to land.

false

Winds always blow from land to water.

false

3. Winds always blow from high to low pressure. How do landmasses impact the movement of air and water?

They stop all movement of both air and water.

They speed up the air and water and allow it to pass though.

They change the direction and slow them down.

think this is it

They slow down the air but speed up the water.

pt:2

answering the pictures

9- I don know sorry

15-B

16-A

Explanation:

just trying to help LET ME KNOW IF IT'S WORNG

3 0
3 years ago
Read 2 more answers
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
emmasim [6.3K]

Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

Thus;

Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

= 298 K

Then ;

Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
The substance which does the dissolving in a solution is
oksano4ka [1.4K]

Answer:

solvents dissolve in liquids

6 0
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