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nydimaria [60]
3 years ago
12

A dialysis unit requires 70000 mL of distilled water. How many gallons of water are needed? (1 gal = 4 qt)

Chemistry
2 answers:
Rasek [7]3 years ago
7 0

75000mL (1 liter/1000 mL)(1 gallon/3.785 liters)  

= [75000*1*1] / [1000*3.785] gallons  = 19.815 gallons

Lady bird [3.3K]3 years ago
5 0

Answer: 19.25 gallons

Explanation: 1 ml = 0.0011 quart

Given:  4 quarts = 1 gallon

Thus if 1 ml is equal to 0.0011 quart

70000 ml  is equal to =\frac{0.0011}{1}\times 70000=77quart

Now if 4 quarts is equal to 1 gallon.

77 quarts is is equal to=\frac{1}{4}\times 77=19.25gallons




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A beam of light passes though a liquid in a test tube without scattering. Which type of mixture is most likely in the test tube?
butalik [34]
The answer would be letter C - solution.

A mixture should be homogeneous for a light not to be scattered. This is because particles are distributed evenly throughout the mixture which allows light to pass directly. In your choices, the solution allows a  beam of light to pass through a liquid in a test tube without scattering.
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3 years ago
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A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi
Aleksandr-060686 [28]

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>

<u><em /></u>

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = <em>9.9652g of water</em>

<em />

I hope it helps!

4 0
3 years ago
Gold forms face-centered cubic crystals. The atomic radius of a gold atom is 144 pm. Consider the face of a unit cell with the n
kipiarov [429]

Answer:

The length of an edge of this unit cell is 407.294 pm

Explanation:

Face centered cubic structure contains 4 atoms in each unit cell and 12 coordination number, occupying about 74% volume of the total cell. Face centered cubic structure is known for efficient use of space for atom packing.

To determine the edge length, a relationship between the radius of the atom and edge length is used.

X = R√8

Where;

X is the length of an edge of this unit cell

R is the radius of the gold atom = 144 pm = 144 X 10⁻¹² m

X = 144 X 10⁻¹²√8

X = 407.294 X 10⁻¹² m

X = 407.294 pm

Therefore, the length of an edge of this unit cell is 407.294 pm

8 0
2 years ago
An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .
grandymaker [24]

Answer: The specific heat of the unknown metal is 0.897J/g^0C

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The specific heat of the unknown metal is 0.897J/g^0C

4 0
3 years ago
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