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3 years ago

9

2. What type of bonding would be expected between the following atoms?

1 answer:

maksim [4K]3 years ago

7 0

A.) covalent
c.) covalent
b.) ionic
d.) covalent

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At what phase of the cell cycle does the genetic material double?

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There are the four phases of the cell cycle. M, G1, S and G3. And in S (Synthesis) phase genetic material is doubled.

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3 years ago

2. What is the current running through a toaster if the resistance is 15

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3 years ago

the combustion of octane is given in this reaction 2C8H18+2502-----16CO2+18H2 O, identify the limit reagent for this reaction wi

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Answer:

Liming reagent in the given reaction is oxygen.

Explanation:

Mass of octane = 12.85 g

Mass of oxygen = 7.46 g

Molar mass of octane = 114.23 g/mol

Molar mass of oxygen = 32

$Mole=\frac{Mass\;in\;g}{Molar\;mass}$

$No.\;of\;moles\;of\;octane=\frac{12.85}{114.23} = 0.1125\;mol$

$No.\;of\;moles\;of\;oxygen=\frac{7.46}{32} = 0.2331\;mol$

$2C_8H_{18} + 25O_2 \rightarrow 16CO_2 +18H_2O$

According to the reaction, 2 moles of octane requires 25 moles of O2

so, 0.1125 moles of octane requires $0.1125\times \frac{25}{2} = 1.40625\;mol$ of oxygen.

but only 0.2331 mol of oxygen is present.

Hence, oxygen is the limiting reagent

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3 years ago

you are assisting in taking a final impression of a patient's mandibular arch, using an automix material. after transferring the

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Explanation:

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3 years ago

The compound 1,1-difluoroethane decomposes at elevated temperatures to give fluoroethylene and hydrogen fluoride: CH3CHF2(g) → C

Maru [420]

Answer : The final temperature would be, 791.1 K

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $460^oC$ = $5.8\times 10^{-6}s^{-1}$

$K_2$ = rate constant at $T_2$ = $4\times K_1$

$Ea$ = activation energy for the reaction = 265 kJ/mol = 265000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $460^oC=273+460=733K$

$T_2$ = final temperature = ?

Now put all the given values in this formula, we get:

$\log (\frac{4\times K_1}{K_1})=\frac{265000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{733K}-\frac{1}{T_2}]$

$T_2=791.1K$

Therefore, the final temperature would be, 791.1 K

4 0

3 years ago