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Crank
2 years ago
8

Which other element is most likely to be a nonreactive gas?

Chemistry
2 answers:
iragen [17]2 years ago
8 0

Option [D] Krypton, Kr number 36

icang [17]2 years ago
4 0

Answer:

Choice D. Among the choices, \rm Kr (krypton) is likely to be the most similar to neon and behave as a non-reactive gas.

Explanation:

Columns in the periodic table are called "groups." Elements in the same group (i.e., same column of the periodic table) tend to have similar chemical properties.

In this example, neon is in the same right-most column of the periodic table. Since neon is a non-reactive gas, it is reasonable to infer that other elements in the same column would behave similarly.

Among the choices, krypton ({\rm Kr}) is the only element in the same column. Thus, krypton would be the most likely element to be a nonreactive gas.

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The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab. How many moles of dextrose is this equivalent t
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The AP Biology teacher is measuring out 638.0 g of dextrose (C6H12O6) for a lab the moles of dextrose is this equivalent to is 3.6888 moles.

<h3>What are moles?</h3>

A mole is described as 6.02214076 × 1023 of a few chemical unit, be it atoms, molecules, ions, or others. The mole is a handy unit to apply due to the tremendous variety of atoms, molecules, or others in any substance.

To calculate molar equivalents for every reagent, divide the moles of that reagent through the moles of the restricting reagent. The calculation is follows:

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1 year ago
Which molecule has a nonpolar covalent bond?
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A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is
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Answer:

\large \boxed{\text{B.) 2.8 atm}}

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ = 1520 Torr; T₁ =   27 °C

p₂ = ?;               T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (  27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\

(c) Convert the pressure to atmospheres

p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}

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