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Sergeu [11.5K]
3 years ago
8

When a patient comes to a hospital with severe dehydration, doctors will give them an IV filled with a salt-water solution calle

d saline, which is isotonic with the body's blood cells. Why is the patient not given pure water? *
Chemistry
1 answer:
kykrilka [37]3 years ago
5 0

Answer:

Patients with dehydration are given salt-water solution or saline water which creates an isotonic environment with blood cells because the isotonic solution restricts the movement of water by creating an equal concentration of solute and water on both sides of the blood membrane and it prevents any change in the cell size.

The patient is not given pure water because it makes the blood cell hypotonic and water will enter the blood cell, and the cell will swell, which can lead to the death of the patient.

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Dolomite, [CaMg(CO3)2] is found in a soil sample. A geochemist titrates 24.65 g of soil with 57.85 mL of 0.3315M HCl. What is th
pshichka [43]

In order to find out the %mass dolomite in the soil, calculate for the mass of dolomite using the information given from the titration procedure. You would need to multiply 57.85 ml with 0.3315 M HCl and you would get the amount of HCl in millimoles. Then multiply the amount of HCl with 1/2 (given that for every 1 mol of dolomite, 2 mol of HCl would be needed). Convert the amount of dolomite to mass by multiplying the millimoles with the molecular weight which is 184.399. Then convert the mass to grams which is 1.768 grams. Divide the mass of dolomite (1.768 grams) with the weight of soil sample. The % mass is 7.17. 

8 0
3 years ago
A vessel of volume 22.4 dm3 contains 2.0 mol h2 and 1.0 mol n2 at 273.15 k initially. all the h2 reacted with sufficient n2 to f
yanalaym [24]
Volume = 22.4 dm3
n = 2 mol of H2
n = 1 mol of N2
Temperature = 273.15
All H2 reacts
reaction
N2 + 3H2 = 2NH3
1:3 ratio
Calculation:
N2 initial - N2 reacted = Final N2
1 - 2*(1/3) = 0.3333 mol of N2 left
H2 = 0 left
NH3 formed = 2/3*1 = 2/3 = 0.666
Total mol:
0.3333 + 0.666 = 1 mol
Apply the equation : 
PV = nRT
P = nRT/V = 1*0.0082*(273.15)/(22.4) = 0.0999924 atm
PH2 = 0
PN2 = 1/3*0.0999924 = 0.0333308 atm
PNH3 = 2/3*0.0999924 = 0.0666616 atm
Answer is 0.0666616 atm
7 0
3 years ago
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate
vovangra [49]

Answer:

K2 = 61.2 M^-1.S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol . If the rate constant of this reaction is 6.7M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

            k = Ae^-(Ea/RT)

where,   k = rate constant

              A = pre-exponential factor

          Ea  = activation energy

             R = gas constant

              T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln(k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

         

      T1= (244 + 273.15) K = 517.15 K

      T2= (324+ 273.15) K =597.15 K

    K1  = 6.7 ,     K2 = ?

         R = 8.314 J/mol K

     Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln(k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^(4.114)

K2 = 61.2

Hence, K2 = 61.2 (M.S)^-1

7 0
3 years ago
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Viefleur [7K]

Answer:

20m/s.

Explanation:

8 0
3 years ago
Becomes caramel. He pushes
kozerog [31]

Answer:

Turning Sugar into Caramel

Explanation:

You turn sugar into caramel by burning sugar. Burning sugar is a chemical change.

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