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statuscvo [17]
3 years ago
13

A. Which beetles is natural selection selecting AGAINST?

Chemistry
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

escarabajos cornudos

Explanation:

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A gas at constant temperature is enclosed in a container with a movable piston. It occupies a space of 10L at a pressure of 500
Liula [17]

Answer:

4.17L

Explanation:

V1 = 10L

V2 =?

P1 = 500torr

P2 = 1200torr

Boyle's law states that at constant temperature, the volume of a gas is inversely proportional to its pressure.

P1V1 = P2V2

V2 = ( P1 * V1 ) / P2

V2 = 4.17L

The new volume of the gas is 4.17L

8 0
3 years ago
Which of the following describes a covalent bond?
Marizza181 [45]

Answer:

D. It is the sharing of electrons between atoms with an electronegativity difference below 1.7

Covalent bonds share electrons, whereas ionic bonds exchange electrons. Covalent bonds have an electronegativity of 0.0-1.7 (0.0-0.3 is a nonpolar covalent bond and 0.3-1.7 is a polar covalent bond). Ionic bonds are bonds that go beyond the electronegativity of 1.7 to 4.0 (1.7-4.0).

7 0
3 years ago
Read 2 more answers
What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
What kind of sport do you have interest the most and explain why? <br>​
lidiya [134]

Answer

Gymnastics

Explanation:

for the simple fact that they do cool stuff

7 0
2 years ago
For which of the following equilibria does `"K"_("eq") = ["O"_2]`? A. O2(l) O2(g) B. 2O3(g) 3O2(g) C. 2H2O(l) 2H2(g) + O2(g) D.
Feliz [49]

For plato users

the answer is a. O2(l) O2(g)

hope this helps!

8 0
3 years ago
Read 2 more answers
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