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kompoz [17]
2 years ago
14

How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?

Chemistry
1 answer:
ser-zykov [4K]2 years ago
5 0

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

10<em>x</em> = 75

<em>x</em> = 75/10 = 7.5

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When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro
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<span>Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
7 0
2 years ago
Use the virtual lab to prepare 150.0 ml of an iodine solution with a concentration of 0.06 g/ ml from the bottle of 0.12g/ml iod
son4ous [18]

Answer:

Explanation:

In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g

Let volume of given concentration of .12 g / ml required be V

In volume V , gram of iodine = V x .12 g

According to question

V x .12 = 9 g

V = 9 / .12 = 75 ml

So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .

8 0
2 years ago
On a separate sheet of paper, write the formulas for the missing components of these neutralization reactions.
Fofino [41]
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H2SO4  +  2NH4OH--------> (NH4)2SO4  +  2H2O

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4 0
2 years ago
Read 2 more answers
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timofeeve [1]

Answer:

K = 6.5 × 10⁻⁶

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Use PV=nRT to find the initial pressure of C₅H₆O₃

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P = 0.78atm

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             = 0.489atm

C₂H₆ = 0.288atm

CO = 0.846atm

K_p = \frac{0.288 * 0.864^3}{0.489}

     = 0.379

K = \frac{K_p}{RT^3}

K = \frac{0.379}{(0.0821 * 473)^3}

   = 6.5 × 10⁻⁶

7 0
2 years ago
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