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Montano1993 [528]
3 years ago
11

In the treatment of shock, which vasoactive drug results in reduced preload and afterload, reducing the oxygen demand of the hea

rt?
Chemistry
1 answer:
USPshnik [31]3 years ago
5 0

Answer:

Nitropruside

Explanation:

Shock is a serious medical condition where oxygen levels in the body are low,  causing a low blood pressure which can lead to organ damage and sometimes death. Shock can be caused by low blood volume or inadequate pumping action of the heart.

Nitropruside is a very potent vasodilator. It acts on the arterial and venous smooth muscles, causing smooth muscle relaxation and leading to decreased cardiac preload and afterload. A reduction in afterload gives rise to an increased cardiac output and blood supply to the cells is increased. This in turn reduces the need for the heart to pump more blood (a compensatory response by the heart due to the low oxygen levels), thereby reducing the oxygen demand of the heart.  

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<span>Movement of water from an area of lower solute concentration to an area of higher solute concentration</span>
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¿Cómo se escribe la fórmula molecular (orden de los iones)?
cluponka [151]

Answer:

A la izquierda el catión y a la derecha el anión.

Explanation:

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En este caso, y basado en las normas IUPAC para la escritura de las fórmulas moleculares, es necesario primero escribir el catión a la izquerda, seguido del anión a la derecha, tal y como se muestra en los siguientes ejemplos, recordando que el catión es el ion cargado positivamente y el anión, negativamente:

K^+Cl^-\\\\Ag_2^+(SO_4)^{2-}

Los cuales son cloruro de potasio y sulfato de plata respectivamente. También es necesario tener en cuenta que los metales tienden a ser cationes por su capacidad de perder electrones, mientras que los no metales a ganarlos y por ende resultar como aniones.

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3 years ago
Balance <br> AI(NO3)3 +<br> H2SO4 →<br> Al2(SO4)3 +<br> HNO3
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Human activities can accelerate the process of
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I believe the answer is C) both

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4 0
3 years ago
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Thepotemich [5.8K]

<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g

  • <u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

Hence, the final temperature of the mixture is  49°C

4 0
3 years ago
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