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sesenic [268]
3 years ago
5

Describe briefly the changes from the control values in the MEAN values for volume, specific gravity and NaCl after drinking pur

e water. Based on the specific gravity measurements, is the urine isosmotic, hyperosmotic or hypoosmotic to the fluids of the body. Explain using physiological relevance.
Chemistry
1 answer:
Yuri [45]3 years ago
8 0

Answer:

As specific gravity is defined as weight per unit volume,so urine is the less dense fluid,so urine will be in hyperosmotic category.

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As a science major, you are assigned to the "Atomic Dorm" at college. The first floor has one suite with one bedroom that has a
lidiya [134]

Explanation:

A.

The first student will be on the lower bunk on the first floor because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available.

B.

7 students are in the TOP bunks because 1. They want on the lowest available floor and 2. They want to be in a lower bunk if available. Therefore, all the rooms up till the third floor (Remember, third floor has 3 suites), so the first floor is filled - 1 person on the top bunk, 2 floor is filled- 4 persons and the third floor; the first suite is filled - 1 person and the second suite is a little partially filled- 1 person.

C.

Following the criteria 1, 2 and 3, the 21st student occupies the third suite on the third floor because all the floors (1 and 2) are occupied so the third suite on the third floor is still vacant.

D.

From the criteria there are therefore 10 persons at the TOP bunk. All the rooms up till the third floor are filled, so the first floor is filled - 1 person on the top bunk, second floor is filled (2 suites) - 4 persons and the third floor; the first suite and second suite is filled - 4 persons; the thirs suite has 6 persons present so 1 person is at the top bunk.

8 0
3 years ago
How does friction affect energy transformation
Westkost [7]
The friction can be used as a stopper for the electricity so it can slow

or it can increace speed
4 0
3 years ago
Read 2 more answers
A patient needs 40.0 mg of antibiotic per kilogram of body weight each day. If the patient weighs 55 kilograms.
White raven [17]

2200 mg of antibiotic

Explanation:

Given that 40 mg of antibiotic/kg of the bodyweight is given.

If patient is 55 kg then  the dose of antibiotic will be

if 40/1000000 is done then we can get antibiotic in kg/kg of the weight

= 0.00004 kg of antibiotic per kg

0.00004*55 ( to know how much 55 kg person will require)

= 0.0022 kg

This 0.0022 value will be converted to mg

0.0022*10^6

= 2200 mg of antibiotic will be given to a 55kg patient.

4 0
3 years ago
22.4l of ammonia is reaxts with 1.406 mole of oxygen to produce NO and h2o .1.what volume of no is produced at ntp​
IceJOKER [234]

Answer:

The volume of NO is 22.4L at STP

Explanation:

Based on the reaction:

2NH3 + 5/2O2 → 2NO + 3H2O

<em>2 moles of NH3 react with 5/2 moles of O2 to produce 2 moles of NO.</em>

<em />

To solve this question, we need to find the moles of each reactant in order to find the limiting reactant as follows:

<em>Moles NH3 -Molar mass: -17.01g/mol-</em>

Using PV = nRT

PV/RT = n

<em>Where P is pressure = 1atm at STP</em>

<em>V is volume = 22.4L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K</em>

1atm*22.4L/0.082atmL/molK*273.15K = n

n = 1.00 moles of NH3

For a complete reaction of 1.00 moles of NH3 are needed:

1.00 moles NH3 * (5/2moles O2 / 2moles NH3) = 1.25 moles of O2

As there are 1.406 moles of O2, <em>the limiting reactant is NH3</em>

<em />

The moles of NO produced are the same than moles of NH3 because 2 moles of NH3 produce 2 moles of NO. The moles of NO are 1.00 moles

And as 1.00moles of gas are 22.4L at STP:

<h3>The volume of NO is 22.4L at STP</h3>

4 0
3 years ago
Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr
zysi [14]

<u>Answer:</u> The value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]

For the given chemical reaction:

N_2+2O_2\rightarrow 2NO_2

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]

We are given:

\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = \frac{121.77}{1}\times 1.90=231.36 J/K

Hence, the value of \Delta S^o for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0
4 years ago
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