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3 years ago

How do new traits enter a population?

1 answer:

4 0

Answer: By the way that population is treated and also by new borns.

Explanation: new borns are given traits based off their parents but also new traits that are developed off of their own. Now if that population has access to resources and a lot of it. Chances are they will become bratty,selfish and things such as.

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A sample of sodium reacts completely with 0.497 kg of chlorine, forming 819 g of sodium chloride. What mass of sodium reacted?

dolphi86 [110]

Answer:

$819 \: g = 819 \times {10}^{ - 3} \: kg = 0.819 \: kg \\ 0.819 \: kg - 0.497 \: kg \\ = 0.322 \: kg$

4 0

3 years ago

2 Use the chart to answer the question. Solubility Curves of Various Salts 180 160 140 120 100 80 KNO3 in g NaCl in g KCl in g A

VMariaS [17]

According to the information in the graph, it can be inferred that the amount of solute that will precipitate out of solution at 20°C is 130 grams.

<h3>How to calculate the amount of solute that precipitates out of solution?</h3>

To calculate the amount of solute that precipitates out of solution we must identify the solute data at 80°C and 20°C and identify the difference as shown below:

Quantity of solute at 80°C: 170 grams.
Quantity of solute at 20°C: 40 grams.

170 grams - 40 grams = 130 grams

According to the above, the amount of solute that will precipitate out of solution due to the change in temperature is 130 grams of KNO3.

Note: This question is incomplete because the graph is missing. Here is the graph

Learn more about solute in: brainly.com/question/7932885

#SPJ1

7 0

2 years ago

The Henry's law constant (kH) for O2 in water at 20°C is 1.28e-3 mol/l atm. How many grams of O2 will dissolve in 3.5 L of H2O t

Sophie [7]

Answer : The mass of $O_2$ dissolved will be, 0.2365 grams

Explanation :

First we have to calculate the concentration of $O_2$ .

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = concentration of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 1.65 atm

$k_H$ = Henry's law constant = $1.28\times 10^{-3}mole/L.atm$

Now put all the given values in the above formula, we get:

$C_{O_2}=(1.28\times 10^{-3}mole/L.atm)\times (1.65atm)$

$C_{O_2}=2.112\times 10^{-3}mole/L$

The concentration of $O_2$ = $2.112\times 10^{-3}mole/L$

Now we have to calculate the moles of $O_2$

$\text{Moles of }O_2=\text{Concentration of }O_2\times \text{volume of solution}$

$\text{Moles of }O_2=(2.112\times 10^{-3}mole/L)\times (3.5L)=7.392\times 10^{-3}mole$

Now we have to calculate the mass of $O_2$

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(7.392\times 10^{-3}mole)\times (32g/mole)=0.2365g$

Therefore, the mass of $O_2$ dissolved will be, 0.2365 grams

4 0

3 years ago

At a certain temperature, 0.900 mol SO 3 is placed in a 2.00 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil

puteri [66]

Answer: The value of $K_c$ is 0.0057

Explanation:

Initial moles of $SO_3$ = 0.900 mole

Volume of container = 2.00 L

Initial concentration of $SO_3=\frac{moles}{volume}=\frac{0.900moles}{2.00L}=0.450M$

equilibrium concentration of $O_2=\frac{moles}{volume}=\frac{0.110mole}{2.00L}=0.055M$ [/tex]

The given balanced equilibrium reaction is,

$2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)$

Initial conc. 0.450 M 0 0

At eqm. conc. (0.450 -2x) M (2x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[O_2][SO_2]^2}{[SO_3]^2}$

$K_c=\frac{x\times (2x)^2}{0.450-2x)^2}$

we are given : x = 0.055

Now put all the given values in this expression, we get :

$K_c=\frac{0.055\times (2\times 0.055)^2}{0.450-2\times 0.055)^2}$

$K_c=0.0057$

Thus the value of the equilibrium constant is 0.0057

3 0

4 years ago

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O)? For example if th

oee [108]

Given :

Mass of oxygen containing carbon monoxide (CO) is 2.666 gram .

To Find :

How many grams of carbon (C) would be present in carbon monoxide (CO) that contains 2.666 grams of oxygen (O) .

Solution :

By law of constant composition , a given chemical compound always contains its component elements in fixed ratio (by mass) and does not depend on its source and method of preparation.

So , volume of solution does not matter .

Moles of oxygen , $n=\dfrac{2.666}{16}=0.167\ mole$ .

Now , molecule of CO contains 1 mole of C .

So , moles of C is also 0.167 mole .

Mass of carbon , $m=12\times 0.167=2\ g$ .

Therefore , mass of carbon is 2 grams .

Hence , this is the required solution .

5 0

3 years ago