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3 years ago

2. Convert 120.02g of FrBiz to moles a. 0.19 moles b. 5.34 moles c. 76,932.82 moles

Chemistry

1 answer:

vampirchik [111]3 years ago

6 0

Answer:

PRACTICE MULTIPLE CHOICE QUESTIONS FROM UNIT 10

1.) A cylinder is filled with 2.00 moles of nitrogen, 3.00 moles of argon, and 5.00 moles of helium. If

the gas mixture is at STP, what is the partial pressure of the argon?

(A) 152 torr (B) 228 torr (C) 380. torr (D) 760. torr

2.) Compared to the average kinetic energy of 1 mole of water at 0 oC, the average kinetic energy

of 1 mole of water at 298 K is

(A) the same, and the number of molecules is the same

(B) the same, but the number of molecules is greater

(D) greater, but the number of molecules is the same

3.) If the pressure on a given mass of gas in a closed system is increased and the temperature

remains constant, the volume of the gas will

(A) decrease (B) increase (C) remain the same

4.) Which gas has approximately the same density as C2H6 at STP?

(A) NO (B) NH3 (C) H2S (D) SO2

5.) At a temperature of 273 K, a 400. milliliter gas sample has a pressure of 760. millimeters of

mercury. If the pressure is changed to 380. millimeters of mercury, at which temperature will

this gas sample have a volume of 551 milliliters?

(A) 100 K (B) 188 K (C) 273 K (D) 546 K

Get the answers to these questions

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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

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a titration of 25.0mL of an x M hcl solution with 0.15M NaOH starts at a burette reading for NaOH of 0.20mL . Thr burette readin

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Burette is a very accurate measuring instrument when adding solutions and has a measurement error of 0.05 mL.
Small volumes of solutions can be transferred from the burette at a controllable rate.
In this instance NaOH is in the burette.
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