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3 years ago

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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expeople1 [14]

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = $\frac{[OH^-][HNO_2]}{[NO_2]}$

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

Where X is reaction coordinate

Replacing in Kb:

1.41x10⁻¹¹ = $\frac{[X][X]}{[0.211 -X]}$

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

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What is the momentum of a 750-kg car traveling at a velocity of a 25 m/s?

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AB + C -----> AC +B

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A weather balloon is filled with helium that occupies a volume of 5.37 104 L at 0.995 atm and 32.0°C. After it is released, it r

svp [43]

Answer:

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Explanation:

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Step 2: Calculate the new volume

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⇒with T1 = the initial temperature = 305.15 K

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⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the decreased temperature : 261.45 K

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