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WINSTONCH [101]
3 years ago
6

The cubic centimeter (cm' or cc) has the same volume as a

Chemistry
1 answer:
snow_lady [41]3 years ago
3 0
The answer is B milliliter
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A soft drink contains 12.1% sucrose (C12H22O11) by mass. What volume of the soft drink solution in milliliters contains 102.5 g
galina1969 [7]

The volume of the soft drink solution in milliliters that contains 102.5 g of sucrose is 11.93mL.

<h3>How to calculate volume?</h3>

The volume of a solution can be calculated by dividing the mass by the density. That is;

Volume = mass/density

According to this question, a soft drink contains 12.1% sucrose (C12H22O11) by mass. This means that the mass of the sucrose is

12.1/100 × 102.5 = 12.40g of sucrose

Volume = 12.40g ÷ 1.04g/mL

Volume = 11.93mL

Therefore, the volume of the soft drink solution in milliliters that contains 102.5 g of sucrose is 11.93mL.

Learn more about volume at: brainly.com/question/1578538

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2 years ago
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Lelu [443]
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3 years ago
Omg I’ve been stressing for the past minutes can someone please help me
Alona [7]

Answer:

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Explanation:

5 0
3 years ago
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The boiling process is considered to be____
Dmitry_Shevchenko [17]

your answer is <u>D</u><u>.</u><u> </u>physical change, because a new substance is not formed.

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3 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
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