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Talja [164]
3 years ago
5

Why might a scientist repeat an experiment if he/she did not make a mistake in the first one? An experiment should be repeated t

o A) increase income for increased work. B) prove how interesting the study was. C) ensure the results are accurate. Eliminate D) eliminate steps and save time.
Chemistry
2 answers:
makvit [3.9K]3 years ago
5 0

The answer is C. To ensure the results are accurate.

WINSTONCH [101]3 years ago
5 0

Answer: The answer is C.

Explanation: An experiment must be repeatable, so doing it multiple times can verify the work is correct.

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Mrs.Jacobs dropped an object from 10 meters she knows it did 50 joules of work how much did it weigh?
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A sugar crystal contains approximately 1.4×1017 sucrose (C12H22O11) molecules. What is its mass in milligrams? Express using 2 s
alekssr [168]

Answer:

From the periodic table:

mass of carbon = 12 grams

mass of hydrogen = 1 grams

mass of oxygen = 16 grams

molar mass of surcose = 12(12) + 22(1) + 11(16) = 342 grams

number of molecules = number of moles x Avogadro's number

number of moles = number of molecules / Avogadro's number

number of moles = (2.2x10^17) / (6.02x10^23) = 3.6544 x 10^-7 moles

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mass = number of moles x molar mass

= 1.7 x 10^17/6.022 x 10^23.

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2 years ago
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Assume that the density and heat of combustion of E85 can be obtained by using 85 % of the values for ethanol and 15 % of the va
Alika [10]

Answer:

87.4 J

Explanation:

The density of the gasoline is 0.70 g/mL, and the density of the ethanol is 0.79 g/mL. The heat combustions (the heat released in a combustion reaction) are 5,400 kJ/mol for gasoline, and 1,370 kJ/mol for ethanol.

For 3.5 L of E85, the volumes of gasoline and ethanol are:

Vgasoline = 0.15 * 3.5 = 0.525 L = 5.25x10⁻⁴ mL

Vethanol= 0.85 * 3.5 = 2.975 L = 2.975x10⁻³ mL

The mass of gasoline and ethanol presented in that sample of E85 is the volume multiplied by the density:

mgasoline = 5.25x10⁻⁴ * 0.70 = 3.675x10⁻⁴ g

methanol = 2.975x10⁻³ * 0.79 = 2.35025x10⁻³ g

The number of moles for each substance is it mass divided by its molar mass. The molar masses are 114 g/mol for gasoline, and 46 g/mol for ethanol:

ngasoline = 3.675x10⁻⁴/114 = 3.224x10⁻⁶ mol

nethanol = 2.35025x10⁻³ /46 = 5.109x10⁻⁵ mol

The energy released is the heat combustion multiplied by the number of moles, so:

Egasoline = 5,400 * 3.224x10⁻⁶ = 0.0174 kJ = 17.4 J

Eethanol = 1,370 * 5.109x10⁻⁵ = 0.07 kJ = 70 J

So, the energy released by the E85 is the sum of the energy released by ethanol and gasoline:

The energy released by E85 = 87.4 J

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