All categories

2 years ago

HELP ANYONE PLEASEE?? TY!!!

Chemistry

1 answer:

ELEN [110]2 years ago

4 0

Answer:

so the order is grass, which gets eaten by the grasshopper, which is eaten by the frog, which gets eaten by the snake. grass wpuld be at top, snake at bottom.

Explanation:

hope this helps

You might be interested in

Is my answer right?

Ahat [919]

Yes the answer is 1s22s22p63s23p64s23d5

7 0

3 years ago

An unknown compound a has the molecular formula c7h14o2. compound a absorbs strongly in the ir at 1700 cm-1. the 1h nmr spectral

scoundrel [369]

The structure of compound A would be solid that is dense enough for antimicrobial form

7 0

3 years ago

Read 2 more answers

---

svet-max [94.6K]

No of protons=No of electrons=15

<h3>Mass no:-No of neutrons+No of protons </h3>

$\\ \tt\longmapsto 15+16=31$

Now

It has atomic no 15 hence its Phosphorus(P)

Electronic configuration:-

$\\ \tt\longmapsto 1s^22s^22p^63s^23p^3$ .

Valency=3

Chloride formula:-

$\\ \tt\longmapsto XCl_3$

or.

$\\ \tt\longmapsto PCl_3$

5 0

3 years ago

Read 2 more answers

How far can he travel in 1 hour 6 minutes

Yakvenalex [24]

I haven't got the answer but I'm probably sure that u can actually look up that answer hoped it help

3 0

3 years ago

Read 2 more answers

A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.

NeTakaya

Answer:

solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 $^{0}\textrm{C}$ can be calculated using the information given.

Let's assume solubility of X in water at 17.0 $^{0}\textrm{C}$ is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

or, y = $\frac{3.96}{36}$ = 0.11

Hence solubility of X in water at 17.0 $^{0}\textrm{C}$ is 0.11 g/mL.

3 0

3 years ago