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kupik [55]
3 years ago
13

For the cells in a human body, an isotonic solution is 0.9% NaCl. If a red blood cell is placed in a 1% NaCl solution, what will

happen to the cell?
Chemistry
1 answer:
igomit [66]3 years ago
6 0

Answer: The red blood cell will shrink due to water loss

Explanation:

Because 1% NaCl solution is slightly more concentrated than its isotonic 0.9% form, the red blood cell will lose its cell containing fluids such as water to the MORE concentrated environment.

This water loss after a prolonged period will result in shrinking of the red blood cell.

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geniusboy [140]

Answer: amount of substance n = cV = 2.0 mol/l x 0.25 l = 0.50 mol

Explanation:

7 0
3 years ago
The burning of fossil fuels will release ____ into the carbon-oxygen cycle.
valkas [14]
The answer is Carbon.
6 0
4 years ago
Which is the correctly balanced equation?
Rudik [331]

Answer:

A) Cl2 + 2NaI = 2NaCl + I2

3 0
3 years ago
Read 2 more answers
Question 1(Multiple Choice Worth 3 points)
Bingel [31]

1) Answer:

moles of oxygen (n) = 1.86 moles

Explanation:

according to  Boyle's law the formula to solve this problem is:

PV=nRt

when P is the pressure which equal 1.25 atm

and V is the volume which equal 37.5 L

n is the number of moles which we need to calculate it

R is constant which equal 0.082

t is the temperature in kelvin

By substitution:

1.25*37.5 = n * 0.082 * 307

So n = 1.86 moles

2) Answer:

the volume of oxygen gas = 34 L

Explanation:

at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4L

So when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that

1.5 moles *22.4 L/ 1 mole = approximately 34 L

3) Answer:

The volume of H2 = 2.29 L

Explanation:

according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1

so to know the number of moles of H2 we will get it for Zn first :

number of moles Zn = mass of Zn / molar mass Zn

                             = 5.98 / 65.39 =0.0914 moles

so number of moles H2 = 0.09 moles

by  substitution in the following formula:

PV = nRT

0.978 * V = 0.09 * 0.082 * 298

so The volume of H2 = 2.29 L

4) Answer:

Volume of O2 = 1.4 L

Explanation:

first we have to balance  the equation:

2Na2O2 +2CO2 → 2Na2CO3 + O2

2 mole CO2 give 1 mole of O2 so the molar ratio is 2:1

at STP 1 mole of gas will equal = 22.4 L

             ??? moles of CO2 = 2.8

n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 moles

so when 1 mole of as = 22.4

              0.0625 moles O2 = ???

Volume of O2 =0.0625 moles * 22.4 L/ 1 mole

                        = 1.4 L

5) Answer:

the initial quantity of sodium metal used = 17.2 gram

Explanation:

at STP 1 mole of gas will equal = 22.4 L

so  moles of H2 equal ?? when 8.40 liters of H2 gas were produced

so moles of H2  = 8.4/22.4 =0.375 moles

and according to the balanced equation the molar ratio between H2 ans Na is 1 : 2

so number of moles for Na = 0.375 *2 = 0.75 moles

to get the initial quantity of sodium metal (mass Na) = number of moles * molar mass

mass Na = 0.75 moles * 23 gm/mole= 17.25

6) Answer:

False

Explanation:

because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature

7) Answer:

22.4 liters

Explanation:

1 mole of gas will equal = 22.4 L

because at the Standard temperature must be 273 K and the standard pressure must be 1 atm

so V = nRT/P

       =1 mole * 0.082 * 273 K / 1 aTm

        = 22.4 liters

4 0
4 years ago
The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
SSSSS [86.1K]

Answer:

There will remain 11.47 grams of Al(OH)3

Explanation:

Step 1: Data given

Mass of sulfuric acid = 35.0 grams

Molar mass sulfuric acid = 98.08 g/mol

Mass of aluminium hydroxide = 30.0 grams

Molar mass of aluminium hydroxide = 78.0 g/mol

Step 2: The balanced equation

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al(OH)3 = 30.0 grams / 78.0 g/mol

Moles Al(OH)3 = 0.385 moles

Moles H2SO4 = 35.0 grams / 98.08 g/mol

Moles H2SO4 = 0.357 moles

Step 4: Calculate the limiting reactant

For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20

H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles

There will remain 0.385 - 0.238 = 0.147 moles

Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams

There will remain 11.47 grams of Al(OH)3

6 0
3 years ago
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