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3 years ago

Analyze the food web. If the frog population decreases, what impact will this likely have on the food web?

Chemistry

2 answers:

sergiy2304 [10]3 years ago

8 0

Answer:

the population of grasshoppers will decrease

Explanation:

a food source is becoming more and more limited thus, its predator will also start to die out due to starvation

Elan Coil [88]3 years ago

5 0

The population of grasshoppers will increase

Since the frogs are decreasing there are less grasshoppers.

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Which element would you expect to be more reactive: phosphorus (P) or fluorine (F)

Studentka2010 [4]

Answer:

Fluorine

Explanation:

Fluorine is the most electronegative element because it has 5 electrons in it's 2P shell.

4 0

2 years ago

Read 2 more answers

The equilibrium of 2H 2 O(g) 2H 2 (g) + O 2 (g) at 2,000 K has a Keq value of 5.31 x 10-10. What is the Keq expression for this

nevsk [136]

Answer:

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Explanation:

For a chemical reaction, equilibrium is a state at which the rate of the forward reaction equals that of the reverse reaction. The equilibrium constant Keq is a parameter characteristic of this state which is expressed as a ratio of the concentration of the products to that of the reactants.

For a hypothetical reaction:

xA + yB ⇄ zC

The equilibrium constant is :

$Keq = \frac{[A]^{x}[B]^{y}}{[C]^{z} }$

The given reaction involves the decomposition of H2O into H2 and O2

$2H_{2}O\rightleftharpoons 2H_{2} + O_{2}$

The equilibrium constant is expressed as :

$Keq = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

Since Keq = 5.31*10^-10

$5.31*10^{-10} = \frac{[]H_{2}]^{2}[O_{2}]}{[H_{2}O]^{2}}$

3 0

3 years ago

Read 2 more answers

Hurry PLEASE HELP!

avanturin [10]

B. 11,540

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Nt=25 g

No=100 g

t1/2=5770 years

$\tt 25=100\dfrac{1}{2}^{T/5770}\\\\\dfrac{1}{4}=\dfrac{1}{2}^{T/5770}\\\\2=T/5770\rightarrowT=11540~years$

7 0

3 years ago

A 40.0 mL sample of 0.18 M HCI is titrated with 0.36 M CoHsNH2. Dctermine the pH at these points: At the beginning (before base

Whitepunk [10]

Answer:

at the beginning:

pH = 0.745

Explanation:

HCl is a strong acid, so:

HCl + H2O → H3O+ + Cl-

0.18 M 0.18 0.18.....equilibrium

before base is added:

∴ [ H3O+ ] ≅ <em>C </em>HCl = 0.18 M

⇒ pH = - Log [ H3O+ ] = - Log ( 0.18 )

⇒ pH = 0.745

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3 years ago

What does the atomic number of each atom represent?

saw5 [17]

It's the number of Protons of that atom..

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2 years ago