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3 years ago

11

When you go down a group, does the tendency for a metal to react (become duller) increase or decrease?

1 answer:

Elan Coil [88]3 years ago

7 0

Increase I'm pretty sure

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The electron cloud of HF is smaller than that of F2, however, HF has a much higher boiling point than F2 has. Which of the follo

mafiozo [28]

Answer:D

Explanation:

The high boiling point of HF is not attributable to the dispersion forces mentioned in the question. In HF, a stronger attraction is in operation, that is hydrogen bonding. This ultimately accounts for the high boiling point and not solely the dispersion model as in F2.

7 0

3 years ago

9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1

WITCHER [35]

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.

3 0

4 years ago

Why is the chemical formula 2C NOT an example of a diatomic molecule?

katen-ka-za [31]

Because it represents 2 atoms of carbon not 2 molecules of carbon

8 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIEST !! HELP ASAP

andre [41]

I am almost positive that the answer is D

6 0

3 years ago

Read 2 more answers

A diesel truck produces 47.5kJ of useful output energy from 125kJ of diesel fuel.what is the truck efficiency?.show the solution

Alex Ar [27]

You have to do a report :
$\frac{E_f}{E_i} = \frac{47,5}{125} = 0,38$

So the truck efficiency is 38%.

7 0

3 years ago