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alina1380 [7]
3 years ago
9

A fish may have hundreds of offspring at a time and only a small number can survive. which characteristics of fish might allow t

hem to survive
Chemistry
1 answer:
mafiozo [28]3 years ago
8 0
Well depends on what type of fish the parent is but i guess you can say:
what they eat
teeth
instincts
color
venom
blending in,etc
You might be interested in
A reaction has ∆H = −356 kJ and ∆S = −36 J/K. Calculate ∆G (kJ) at 25°C.
Cloud [144]

Answer: -345.2 KJ

Explanation: As we know that ,dG=dH-TdS

T=25+273=298 K

dG= -356 x1000-298(-36)= -356000+10728

=-345272 j

= -345.2 KJ

5 0
3 years ago
Question 1. Considering the reaction of the decomposition of N2O5 to form the products NO2 and O2, calculate the average reactio
GalinKa [24]
<span>
 </span><span> average reaction rate </span><span>= change in concentration / change in time
by putting values we have

 = (1.00M - 0.987M) / (4.00s - 0.00s)
 = 3.25x10^-3 mol/Lsec
this is our conclusion
hope this helps</span>
7 0
3 years ago
Read 2 more answers
How much<br>much hydrogen gas evolved<br>when 1.5 current is passed through water for 1.5 hours?​
evablogger [386]

0.042 moles of Hydrogen evolved

<h3>Further explanation</h3>

Given

I = 1.5 A

t = 1.5 hr = 5400 s

Required

Number of Hydrogen evolved

Solution

Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.

Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)

Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻

Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)

So at the cathode H₂ gas is produced

Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C

\tt mol~e^-=\dfrac{Q}{96500}

Q = i.t

Q = 1.5 x 5400

Q = 8100 C

mol e⁻ = 8100 : 96500 = 0.084

From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042

4 0
3 years ago
You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
miss Akunina [59]

Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

5 0
3 years ago
100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
5 0
3 years ago
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