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ira [324]
2 years ago
12

____________ acceleration involves speeding up, while ____________ acceleration involves slowing down. a. positive, negative c.

positive, braking b. negative, positive d. both a and c
Chemistry
2 answers:
Novosadov [1.4K]2 years ago
8 0

Answer:

Positive acceleration involves speeding up, while negative acceleration involves slowing down.

Explanation:

I am so sorry if I was wrong!

Fofino [41]2 years ago
5 0

Answer:

no just no

Explanation:

You might be interested in
Calculate the pOH of an aqueous solution of .0.073 M LiOH
Novosadov [1.4K]

Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14

The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).

pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:

pOH= - log [OH⁻]

On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.

LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:

[LiOH]= [OH-]= 0.073 M

Replacing in the definition of pOH:

pOH= -log (0.073 M)

<u><em>pOH= 1.14 </em></u>

In summary, the pOH of the aqueous solution is 1.14

Learn more:

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  • <u>brainly.com/question/13557815?referrer=searchResults</u>
7 0
3 years ago
You drop an unknown substance that weighs 8.3g into a graduated cylinder with 6ml of water. The water rises to 8ml when you drop
valkas [14]

Answer:

<h3>The answer is 4.15 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of object = 8.3 g

volume = final volume of water - initial volume of water

volume = 8 - 6 = 2 mL

So we have

density =  \frac{8.3}{2}  \\

We have the final answer as

<h3>4.15 g/mL</h3>

Hope this helps you

7 0
3 years ago
Describe what golgi looks like.
Margaret [11]
Here is what Golgi looks like, so that you can look at the picture and describe it.
______________________________________________________________
____________________________________________________________________________________________________________________________

Glad I could help, and good luck!
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7 0
2 years ago
The decomposition of a compound at 400⁰C is first order with half-life of 1570 seconds. what fraction of an initial amount of th
kirill115 [55]

Answer: After 4710 seconds, 1/8 of the compound will be left

Explanation:

Using the formulae

Nt/No = (1/2)^t/t1/2

Where

N= amount of the compound  present at time t

No= amount of compound present at time t=0

t= time taken for N molecules of the compound to remain = 4710 seconds

t1/2 = half-life of compound  = 1570 seconds

Plugging in the values, we have  

Nt/No = (1/2)^(4710s/1570s)

Nt/No = (1/2)^3

Nt/No= 1/8

Therefore after 4710 seconds, 1/8 molecules of the compound will be left

5 0
2 years ago
Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kin
Dennis_Churaev [7]

Answer:

35.75 days

Explanation:

From the given information:

For first-order kinetics, the rate law can be expressed as:

\mathsf{In \dfrac{C}{C_o} = -kt}

Given that:

the rate degradation constant = 0.12 / day

current concentration C = 0.05 mg/L

initial concentration C₀ = 3.65 mg/L

\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}

㏑(0.01369863014) = -(0.12) t

-4.29 = -(0.12)

t = -4.29/-0.12

t = 35.75 days

3 0
3 years ago
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