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2 years ago

6

What element in the second period has the largest atomic radius?

1 answer:

maksim [4K]2 years ago

7 0

Answer:

Be

Explanation:

Mg is in the 3rd period. So eliminate it.

Across a period the atomic radius decrease. So Be > C > O

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Why is it important to pasteurize certain liquids?

galben [10]

Answer:

Pasteurization involves heating liquids at high temperatures for short amounts of time, it can make it better if you wanted hot cocoa but bad for milk

Explanation:

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What effect does a quadrupling of the mass have upon the acceleration of the object

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The state in which all of the external forces acting upon an object are balanced; there is no acceleration. friction ..... quadrupling. doubling distance and quadrupling mass has the overall effect of the force

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3 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

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4 years ago

Tarnish on copper is the compound CuO. A tarnished copper plate is placed in an aluminum pan of boiling water. When enough salt

oksian1 [2.3K]

Answer:

The standard cell potential is 2.00 V

Explanation:

<u>Step 1:</u> Data given

Cu is cathode because of higher EP

Al3++3e−→Al E∘=−1.66 V anode

Cu2++2e−→Cu E∘=0.340 V cathode

<u>Step 2:</u> Balance both equations

2*(Al → Al3+-3e−) E∘=1.66 V

3*(Cu2++2e−→Cu) E∘=0.340 V

<u>Step 3:</u> The netto equation

2 Al + 3Cu2+ +6e- → 2Al3+ + 3Cu -6e-

2 Al + 3Cu2+ → 2Al3+ + 3Cu

<u>Step 4:</u> Calculate the standard cell potential

E∘cell = E∘cathode - E∘anode

E∘cell = E∘ Cu2+/Cu - E∘ Al3+/Al

E∘cell =0.340 V - (-1.66) = 2.00 V

The standard cell potential is 2.00 V

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3 years ago

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dusya [7]

Answer:

I think it is b.

Explanation:

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