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3 years ago

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How many moles of cesium xenon heptafloride can be produced from the reaction 12 mol cesium Floride with 14 mol xenon hexaflorid

e

1 answer:

neonofarm [45]3 years ago

7 0

Answer:

12 moles of cesium xenon heptafluoride

Explanation:

The reaction of cesium fluoride with xenon hexafluoride is CeF + XeF6 -> CeXeF7 and the reaction is balanced as written. So the mole ratio is 1:1:1. We are given 12 moles of CeF and 14 moles of XeF6 are reacting, but after the 12 moles of CeF react completely, the reaction will stop as we have run out of one of our reactants. So only 12 moles of CeXeF7 will be produced.

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Carbon-14 has a half-life of 5720 years and this is a first-order reaction. If a piece of wood has converted 88.5% of the carbon

Ivahew [28]

<u>Answer:</u> The sample of Carbon-14 isotope is 37056.3 years old

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5720 years

Putting values in above equation, we get:

$k=\frac{0.693}{5720yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 88.5) = 11.5 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{11.5}\\\\t=37056.3yrs$

Hence, the sample of Carbon-14 isotope is 37056.3 years old

6 0

3 years ago

3. Classify the following plants along with two main characteristics: al Cycus b) Bamboo c) Lemna d) Paddy a) Sugarcane f) Pinus

madreJ [45]

Answer:

A. cycas

1. it is thick and scaly

2. it grow relatively slowly and have a large, terminal rosette of leaves.

B. bamboo

1. it is very durable

2. it is both flexible and elastic

C. lemna

1. it grows as simple free-floating thalli on or just beneath the water surface

2. it are small, not exceeding 5 mm in length

D. paddy

1. it is variety purity

2. it degree of purity

E. sugarcane

1. it is bear long sword-shaped leaves

2. it's stalks are composed of many segments, and in each joint there is a bud

4 0

2 years ago

How many grams are in 1.2 moles of neon

DerKrebs [107]

1.2mole•20.17g/1mole= 24.20g

7 0

3 years ago

Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos

SCORPION-xisa [38]

Explanation:

More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a) Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, $PF_{3}$ is the most ionic compound and $PBr_{3}$ is the least ionic compound.

(b) Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, $BF_{3}$ is the most ionic compound and $NF_{3}$ is the least ionic compound.

(c) Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, $TeF_{4}$ is the most ionic compound and $BrF_{3}$ is the least ionic compound.

6 0

3 years ago

Need help !!!!! ASAP

Nat2105 [25]

<h2>Hello!</h2>

The answer is:

The temperature will be the same, 37°C.

<h2>Why?</h2>

Since from the statemet we know the first temperature, pressure and volumen of a gas, and we need to calculate the new temperature after the pressure and the volume changed, we need to use the Combined Gas Law.

The Combined Gas Law establishes a relationship between the temperature, the pressure and the volume of an ideal gas using Boyle's Law, Gay-Lussac's Law and Charles's Law.

The law establishes the following equation:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}$

Where,

$P_{1}$ is the first pressure.

$V_{1}$ is the first volume.

$T_{1}$ is the first temperature.

$P_{2}$ is the second pressure.

$V_{2}$ is the second volume.

$T_{2}$ is the second temperature.

Then, we are given the following information:

$V_{1}=200mL\\P_{1}=4atm\\T_{1}=37\°C\\V_{2}=400mL\\P_{2}=2atm$

So, isolating the new temperature and substituting the given information, we have:

$\frac{P_{1}V{1}}{T_{1}}=\frac{P_{2}V{2}}{T_{2}}\\\\T_{2}=P_{2}V{2}*\frac{T_{1}}{P_{1}V_{1}} \\\\T_{2}=2atm*400mL*\frac{37\°C}{4atm*200mL}=37\°C$

Hence, we have that the temperature will not change because both pressure and volume decreased and increased proportionally, creating the same relationship that we had before the experiment started.

The temperature will be the same, 37°C

Have a nice day!

4 0

3 years ago