Ask question

Ask question

All categories

3 years ago

6

When bonds are (broken/formed) there is a positive energy change.

2 answers:

valina [46]3 years ago

8 0

Very true, it goes from being positive to becoming negative.

krek1111 [17]3 years ago

3 0

Answer: Hello i am confused are you asking a question?

Explanation:

You might be interested in

Is plasma a long lasting state? Comment on the lifetime of plasma.

Alex787 [66]

Answer:

The correct response will be "Yes". The further explanation is given below.

Explanation:

Plasma seems to be a physical phenomenon where another gasification is energized because once atomic particles were never longer compatible through any terms of direction atom.
They have been completely fabricated of charged particles (+) and unconstrained electrons, which seems to be very similar. Plasma must always be generated and preserved onto Terra.

So that the above is the correct approach.

3 0

4 years ago

SOMEONE HELP NO LINKS PLZ answer Questions 1–6 based on Figure 1.

Bogdan [553]

Answer:

element argon, protons 18 , electrons 18 , outer most energy level 8

6 0

3 years ago

Please help thank you (15 points)

sashaice [31]

Answer:

If the answer helps you PLEASE mark me as brainliest

Stay Safe, Stay Happy And stay healthy

6 0

3 years ago

Which substance can not be decomposed by a chemical change?<br> (1) AlCl3 (3) HI<br> (2) H2O (4) Cu

user100 [1]

Cooper is the substance that can't be decomposed by a chemical chAnge. So 4)

5 0

3 years ago

Read 2 more answers

Given the following reaction: 2N2O5=2N2O4+O2, if the rate of oxygen production is 0.15M/min, determine:

astraxan [27]

Answer:

a. $r_{N_2O_5}=-0.075M/min$

b. $r_{N_2O_4}=0.075M/min$

Explanation:

Hello.

In this case, according to the balanced chemical reaction, we can write the law of rate proportions:

$\frac{r_{N_2O_5}}{-2} =\frac{r_{N_2O_4}}{2} =\frac{r_{O_2}}{1}$

Thus, we proceed as follows:

a. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

$\frac{r_{N_2O_5}}{-2} =\frac{r_{O_2}}{1}\\\\r_{N_2O_5}=\frac{r_{O_2}}{-2} =\frac{0.15M/min}{-2}\\\\ r_{N_2O_5}=-0.075M/min$

b. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

$\frac{r_{N_2O_4}}{2} =\frac{r_{O_2}}{1}\\\\r_{N_2O_4}=\frac{r_{O_2}}{2} =\frac{0.15M/min}{2}\\\\ r_{N_2O_4}=0.075M/min$

Best regards!

8 0

3 years ago