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dolphi86 [110]
3 years ago
7

What is the correct clases for this drugs Promethazine heelp me please

Chemistry
1 answer:
rosijanka [135]3 years ago
5 0

Answer:

Promethazine is in a class of medications called phenothiazines. It works by blocking the action of a certain natural substance in the body.

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The sun melts a popsicle. is what type pf transfer
denis23 [38]

Answer:

heat?

Explanation:

the sun warms the popsicle, therefore melting th popsicle.

sry if this is wrong

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PLEASE PLEASE PLEASE HELP ME HAS TO BE TURNED IN AT 9:00
Kay [80]

Answer:

Option D will be the answer.

8 0
3 years ago
True or false : sound waves travel as particles of matter collide and transfer energy
dimaraw [331]

Answer:

the answer is true

Explanation:

as sound travels through the air it creates a disturbance in the particles of the medium

8 0
3 years ago
An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0100 M O2. At this temperature, Kc
Naya [18.7K]
The correct answer to this question is letter "B) 8.4 × 10-4 M."

2CO+O2<->2CO2

-x       -x        +x

It started as 2CO2 so +x, then it decomposed into 2CO and O2, so-x

2CO-oxidized, O2-Oxidized, 2CO2-reduced


the easiest way is to look at the equation, so if we have some reactant and no product 

A<->2C+D

-x    +2x +x
so if we had product concentration, and no reactant concentration 3A<-> C+D +3x -x-x <span>
</span>
8 0
4 years ago
Read 2 more answers
The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use EscE 0.241V.
tino4ka555 [31]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The  value is   K_b  =   1.89 *10^{-6}

Explanation:

From the question we are told that

   The  voltage of the cell is  V  =  0.731 \  V

Generally K_b is mathematically represented as  

           K_b  =  \frac{K_w }{ K_a }

Where  K_w  is the equilibrium constant for this auto-ionization of water with a value  K_w  =  1.0 *10^{-14}

Generally the E_{cell} is mathematically represented as

       E_{cell} =  V  -  E_{SCE}

=>     E_{cell} =  0.731 - 0.241

=>       E_{cell} =   0.49 V

This  E_{cell} is mathematically represented as

             E_{cell} =  \frac{0.0592}{n} *  log K_a

Where n is the number of moles which in this question is  n = 1

         So  

         0.490 =  \frac{0.0592}{1}  *  log K_a

=>      K_a  =  5.30*10^{-9}

So  

     K_b  =  \frac{ K_w}{ K_a}

=>   K_b  =  \frac{1.0 *10^{-14}}{ 5.30*10^{-9}}

=>    K_b  =   1.89 *10^{-6}

5 0
3 years ago
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