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2 years ago

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Calculate the enthalpy of combustion, δh∘comb, for c6h14. you'll first need to determine the balanced chemical equation for the

combustion of c6h14. express your answer to four significant figures and include the appropriate units.

1 answer:

Vedmedyk [2.9K]2 years ago

3 0

<span>Answer: Enthalpy Change = (6 x -393.5) + (7 x -285.8) - (-204.6) + (19/2) 0.....??? like.. (6 x Enth CO2) + ( 7 x Enth H2O) - (Enth C6H14) + (19/2) Enth O2</span>

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Small beads of iridium-192 are sealed in a plastic tube and inserted through a needle into breast tumors. If an Ir-192 sample ha

Stells [14]

Answer:

296.1 day.

Explanation:

The decay of radioactive elements obeys first-order kinetics.

For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(74.0 days) = 9.365 x 10⁻³ day⁻¹.

For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 9.365 x 10⁻³ day⁻¹).

t is the time of the reaction (t = ??? day).

a is the initial concentration of Ir-192 (a = 560.0 dpm).

(a-x) is the remaining concentration of Ir-192 (a -x = 35.0 dpm).

<em>∴ kt = lna/(a-x)</em>

(9.365 x 10⁻³ day⁻¹)(t) = ln(560.0 dpm)/(35.0 dpm).

(9.365 x 10⁻³ day⁻¹)(t) = 2.773.

<em>∴ t </em>= (2.773)/(9.365 x 10⁻³ day⁻¹) =<em> 296.1 day.</em>

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2 years ago

In a 0.01 M solution of HCl, Litmus will be

Nadya [2.5K]

In a 0.01 M solution of HCl, Litmus will be red. Litmus paper will turn into red in acidic conditions. Hydrochloric acid is an acid. Litmus is an indicator for acidity and alkalinity made from inchens.

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2 years ago

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Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Virty [35]

Thee question is incomplete; the complete question is;

Which elements do not strictly follow the octet rule when they appear in the Lewis structure of a molecule?

Select one or more:

A: Chlorine

B: Carbon

C: Hydrogen

D: Sulfur

E: Fluorine

F: Oxygen

Answer:

chlorine

sulphur

Explanation:

The octet rule states that, for atoms to be stable, they must have eight electrons on their outermost shells.

This rule is not strictly followed by some elements such as sulphur and chlorine. The atoms of these elements can sometimes expand their octet by utilizing the d-orbitals found in the third principal energy level and beyond.

These leads to formation of compounds in which the central atom has more than eight electrons in its outermost shell.

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2 years ago

A 0.380 kg sample of aluminum (with a specific heat of 910.0 J/(kg x K)) is heated to 378 K and then placed in 2.40 kg of water

lbvjy [14]

Answer:

The equilibrium temperature of the system is 276.494 Kelvin.

Explanation:

Let consider the system formed by the sample of aluminium and water as a control mass, in which the sample is cooled and water is heated until thermal equilibrium is reached. The energy process is represented by First Law of Thermodynamics:

$Q_{water} -Q_{sample} = 0$

$Q_{water} = Q_{sample}$

Where:

$Q_{water}$ - Heat received by water, measured in joules.

$Q_{sample}$ - Heat released by the sample of aluminium, measured in joules.

Given that no mass is evaporated, the previous expression is expanded to:

$m_{w}\cdot c_{p,w}\cdot (T-T_{w}) = m_{s}\cdot c_{p,s}\cdot (T_{s}-T)$

Where:

$m_{s}$ , $m_{w}$ - Mass of water and the sample of aluminium, measured in kilograms.

$c_{p,s}$ , $c_{p,w}$ - Specific heats of the sample of aluminium and water, measured in joules per kilogram-Kelvin.

$T_{s}$ , $T_{w}$ - Initial temperatures of the sample of aluminium and water, measured in Kelvin.

$T$ - Temperature which system reaches thermal equilibrium, measured in Kelvin.

The final temperature is now cleared:

$(m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s})\cdot T = m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}$

$T = \frac{m_{s}\cdot c_{p,s}\cdot T_{s}+m_{w}\cdot c_{p,w}\cdot T_{w}}{m_{w}\cdot c_{p,w}+m_{s}\cdot c_{p,s}}$

Given that $m_{s} = 0.380\,kg$ , $m_{w} = 2.40\,kg$ , $c_{p,s} = 910\,\frac{J}{kg\cdot K}$ , $c_{p,w} = 4186\,\frac{J}{kg\cdot K}$ , $T_{s} = 378\,K$ and $T_{w} = 273\,K$ , the final temperature of the system is:

$T = \frac{(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)\cdot (378\,K)+(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)\cdot (273\,K)}{(2.40\,kg)\cdot \left(4186\,\frac{J}{kg\cdot K} \right)+(0.380\,kg)\cdot \left(910\,\frac{J}{kg\cdot K} \right)}$

$T = 276.494\,K$

The equilibrium temperature of the system is 276.494 Kelvin.

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2 years ago

Write a brief passage describing a neutral atom of nitrogen-14 (N-14). Describe the number of protons, neutrons, and electrons i

Inessa [10]

In nitrogen-14, there are 7 protons, 7 neutrons, and 7 electrons. The protons and neutrons are in the nucleus, and the electrons are in the electron shells. The atomic number is the number of protons, the mass number is the number of protons AND neutrons, and the atomic mass is the average of the masses of all isotopes.

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2 years ago

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