Ask question

Ask question

All categories

3 years ago

7

In what way does the shape of a molecule affect how the molecule is

1 answer:

AlexFokin [52]3 years ago

5 0

Answer:

It determines how the molecule functions.- B.

You might be interested in

Which of the following observations would indicate that a material is a pure substance

Lena [83]

What is the following

3 0

2 years ago

Read 2 more answers

How does water affect the growth of a plant?

irina1246 [14]

Answer:

Water transports important nutrients throughout the plant.

Explanation:

Much like blood in an animal, water is necessary for a plants survival, as it transports important nutrients from the soil all through the plant's stems and leaves.

3 0

3 years ago

Read 2 more answers

The article explains that elephants sometimes communicate in low ultrasound. Based on what you learned above, would infrasound w

Likurg_2 [28]

I have not read the article, and so I don't know in what context this is referring to. But I do know that <span>Infra-sound is made up of a really low frequency sound, beyond the range of hearing by humans. </span>Helpful? :)

6 0

3 years ago

Kevin has 5 fish in his fish tank jasmine has 4 times as many fish as Kevin.How many fish does jasmine have?

madreJ [45]

Jasmine have 20 fish because4 times 5 equals 20

4 0

3 years ago

The activation energy (E*) for 2N2O ---> 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be

Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature $T_1 =$ 1001 K

Final temperature $T_2 =$ 298 K

Applying the equation of Arrhenius theory.

$In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})$

where ;

R gas constant = 8.314 J/K/mol

$In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})$

$In \dfrac{k_2}{0.380 }= -70.8655$

$\dfrac{k_2}{0.380 }= e^{-70.8655}$

$\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}$

${k_2}= 1.67303256 \times 10^{-31} \times {0.380 }$

${k_2}= 6.3575 \times 10^{-32}$ /M .sec

Half life:

At 1001 K.

$t_{1/2} = \dfrac{In_2}{k_1}$

$t_{1/2} = \dfrac{0.693}{0.38}$

$t_{1/2} =$ 1.82368 secc

At 298 K:

$t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}$

$t_{1/2} =1.09 \times 10^{31} \ sec$

3 0

3 years ago