All categories

3 years ago

24 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry

2 answers:

erastovalidia [21]3 years ago

6 0

I'd say b, but i'm not 100 percent sure.<span />

Oksana_A [137]3 years ago

6 0

I would say D, because it did not say you are just starting the flight. If it did, then it would be B, because you need to speed up and you are going upwards. It said you are going in the same amount of speed, and of course, if you are going in the same speed, you can't make time pass faster or slower. So, if it was me, I would choose D as an answer.

You might be interested in

Need help on this fast

soldi70 [24.7K]

Salt is the solute, water is the solvent

3 0

3 years ago

Read 2 more answers

If an atom of the following element had a neutral charge, how many electrons would it have?

Yakvenalex [24]

Answer:

Explanation:

In order for an atom to have a neutral charge, it needs to have the same number of protons and electrons. You can see by the illustration that the atomic number is 31, so this element has 31 protons. Knowing that, we also know that it must have 31 electrons to be neutral.

8 0

2 years ago

A mass spectrometer has determined the mass and abundances of all isotopes of an unknown element. The first isotope has a mass o

shepuryov [24]

Answer:

I believe it is Potassium (K)

Explanation:

I did the math on a calculator and it was the closest atomic mass to potassium.

5 0

3 years ago

Help me please I really do need help pls text me if u wanna help me more

8090 [49]

Respiration . I’m pretty sure

8 0

3 years ago

Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....

Oduvanchick [21]

Answer:

Explanation:

$\mathbf{From \ the \ information \ given:} \\ \\ \mathbf{The \ rates \ of \ the \ f ollowing \ reactions \ can \ be \ expressed \ as \ follows:}$

(a)

$\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}$

(b)

$\mathbf{C_2H_6 \to C_2H_4 + H_2} \\ \\ \\ \mathbf{ -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}$

(c)

$\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} = \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt} }$

(d)

$\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\ \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}$

(e)

$\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\ \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}$

5 0

2 years ago