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anyanavicka [17]
3 years ago
13

Xavier was conducting a long term study of two populations of butterflies. Both populations had the same number of individuals.

Both populations lived in forests that were the same size. Population A lived in a forest that was full of milkweed, the favorite food of this butterfly species. Population B lived in a forest that had limited amounts of milkweed and other foods. Over many years, Xavier found that one of the populations was showing signs of evolutionary change. Their mouth parts were changing and they could get food from new kinds of plants. Which statement explains why one population would be more likely to evolve in response to natural selection? *
a. Population B would be more likely to change because it has a ready supply of food, and only certain butterflies would be able to survive
b. Population A would be more likely to change because it has a ready supply of food and only certain butterflies would be able to survive
c. Population A would be more likely to change because it has a limited amount of food and some butterflies might be better able to get enough food, survive, and reproduce
d. Population B would be more likely to change because those butterflies have a limited supply of food, and butterflies with certain traits might be better able to get enough food, survive, and reproduce.
Chemistry
1 answer:
nata0808 [166]3 years ago
8 0
B. I had that same question!!
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What is the engine piston displacement in liters of an engine whose displacement is listed as 490 in^3?
marishachu [46]

Answer:

490 in^3 = 8.03 L

Explanation:

Given:

The engine displacement = 490 in^3

= 490 in³

To determine the engine piston displacement in liters L;

(NOTE: Both in^3 (in³) and L are units of volume). Hence, to find the engine piston displacement in liters (L), we will convert in^3 to liters (L)

First, we will convert in³ to cm³

Since 1 in = 2.54 cm

∴ 1 in³ = 16.387 cm³

If 1 in³ = 16.387 cm³

Then 490 in³ =  (490 in³ × 16.387 cm³) / 1 in³ = 8029.63 cm³

∴ 490 in³ = 8029.63 cm³

Now will convert cm³ to dm³

(NOTE: 1 L = 1 dm³)

1 cm = 1 × 10⁻² m = 1 × 10⁻¹ dm

∴ 1 cm³ = 1 × 10⁻⁶ m³ = 1 × 10⁻³ dm³

If 1 cm³ = 1 × 10⁻³ dm³

Then, 8029.63 cm³ = (8029.63 cm³ × 1 × 10⁻³ dm³) / 1 cm³ = 8.02963 dm³

≅ 8.03 dm³

∴ 8029.63 cm³ = 8.03 dm³

Hence, 490 in³ = 8029.63 cm³ = 8.03 dm³

Since 1L = 1 dm³

∴ 8.03 dm³ = 8.03 L

Hence, 490 in³ = 8.03 L

3 0
3 years ago
How many liters of h2 gas, collected over water at an atmospheric pressure of 752 mm hg and a temperature of 21.0°c, can be made
natta225 [31]
Answer:  
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.  
1.566 g Zn x (1 mole Zn / 65.38 g Zn) = 0.02395 moles Zn  
0.02395 moles Zn x (1 mole H2 / 1 mole Zn) = 0.02395 moles H2 produced  
Now use the ideal gas law to find the volume V.  
P = 733 mmHg x (1 atm / 760 atm) = 0.964 atm  
T = 21 C + 273 = 294 K  
PV = nRT 
V = nRT/ P = (0.02395 moles H2)(0.0821 L atm / K mole)(294 K) / (0.964 atm) = 0.600 L
7 0
3 years ago
Select a system in the human body and explain how a breakdown in the digestive system might affect the system you just chose.\
Citrus2011 [14]
The circulatory system works with the digestive system. Once the food is digested, the circulatory systems absorb and uses the nutrients in the food. If the digestive system were to break down, the circulatory will not have the nutrients it needs to sufficiently run the body. 
6 0
3 years ago
Lewis structure for 2O2
Mrrafil [7]

the big number describes the number ratio in a chemical equation

so for example,

2H2 + O2 --> 2H2O means

2 moles of hydrogen reacts with one mole of oxygen to form 2 moles of water

and as you know, the small (subscript) number determines the number of atoms of that element in one molecule of a compound

so I believe that drawing a normal lewis structure ( O=O ) should be correct

6 0
3 years ago
Read 2 more answers
S8 + 24 F2 ⟶ 8 SF6
Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

= 425g/256g/mol.      = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant. </em>

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

<em>Determining Theoretical Yield:</em>

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.  

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

   = 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈ </em>

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆  to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)  

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.        

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.    

____________________

Here's another example just for grins ...

             C₂H₆O   +   3O₂     =>     2CO₂    + 3H₂O

Given:    253g          307g               ?               ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced        

Limiting Reactant

moles  C₂H₆O = 253g/46g/mol = 5.5 moles  => 5.5/1 = 5.5

moles  O₂ = 307g/32g/mol = 9.6 moles         =><em>  9.6/24 = 0.4 ∴ O₂ is L.R.</em>

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.  

 C₂H₆O   +       3O₂          =>     2CO₂    + 3H₂O

------------ 9.6 mole (L.R.)              ?               ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole  (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole  = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient  C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

7 0
3 years ago
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