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Leni [432]
3 years ago
8

Assume that a milliliter of water contains 20 drops. How long, in hours, will it take you to count the number of drops

Chemistry
1 answer:
garri49 [273]3 years ago
6 0

Answer:

126.18 hr

Explanation:

Data given:

1 mL of water = 20 drops

count rate = 10 drops/s

time in hours for one gallon = ?

Solution:

First we calculate number of mL (milliliter) of water in gallon

As we know

1 gallon = 3785.4 mL

As,

1 galon consist of 3785.4 mL of water, so now we count number of drops that contain 3785.4 mL of water

As we Know 1 mL water contain 20 drops then 3785.4 mL of water contain how many drops:

Apply unity formula

                    1 mL water ≅ 20 drops

                    3785.4 mL water ≅ X drops

Do cross multiplication

                 X drops of water = 20 drops x 3785.4 mL / 1 mL

                 X drops of water = 75708 drops

So, we come to know that one gallon contain 75708 drops of water and we have to calculate the time in hour to count these drops

First we calculate time in seconds

As we Know 10 drops water count in one second then how many seconds it will take to count 75708 drops

Apply unity formula

                    1 second ≅ 10 drops

                    X second ≅ 75708 drops

Do cross multiplication

                 X second  = 1 second x 75708 drops / 10 drops

                 X second = 7570.8 second

So it take 7570.8 second to count 1 gallon water drops

Now convert seconds to hours

As,

60 seconds = 1 hr

7570.8 second  =  7570.8 / 60 = 126.18 hr

So it take 126.18 hr to count 1 gallon water drops.

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Answer:

183 cg = 0.00183 kg

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Explanation:

Use conversion factors. 1kg is equal to 1 x 10^5 cg (100000) and 1 kg is equal to 1 x 10^3 grams (1000 grams).

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Which out of two has stronger forces of attraction sugar or water very sort answer​
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Liquids and solids are referred to as "condensed phases" because the attractive forces cause molecules to become stationary. lea
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leave little space between the molecules.

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A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9
Ronch [10]

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂)  = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺)   = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻)    = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

      = -260.75 + 232.34

     = -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

   = 1/(0.750 * 0.232²)

  = 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

     = -28.41 +7.95

    = -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

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