Who among the following is most likely to have elevated levels of endorphins

Earths climate history <br> Short writing

ra1l [238]

Answer:

Earth's climate has changed throughout history. Just in the last 650,000 years there have been seven cycles of glacial advance and retreat, with the abrupt end of the last ice age about 11,700 years ago marking the beginning of the modern climate era — and of human civilization. Most of these climate changes are attributed to very small variations in Earth’s orbit that change the amount of solar energy our planet receives.

Explanation:

In short words , Earths climate went from cold this is due because ; of the global Ice Age. After the rock showers hit it released large amounts of carbon it heated up the atmosphere causing the word to have bacteria and grow livable life-forms again. 2021 we are producing too much carbon and this is leading factor of climate change not only is the earth already warm enough but its rapidly heating up. An example of this is ETC - "I live in CA and since the start of Autumn we haven't had and glimpse or foreshadowing of cold weather its been 80* everyday and hasn't dropped one bit not even in the morning."

4 0

3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

OleMash [197]

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

4 0

4 years ago

There are 15 technicians and 11 chemists working in a research laboratory. In how many ways could they form a 5-member safety co

Fynjy0 [20]

a)

26C5 -> 26 choices of 5 things

b)

15C1 × 11C4

c)

11C1 × 15C4

d)

11C2 × 15C2

e)

11C4

or

15C4

5 0

3 years ago

. The density of gold is 19.32 g/mL. A U.S. $5.00 gold coin has a diameter of 1.62 cm and a thickness of 0.085 cm. The price of

Lostsunrise [7]

Answer:

$182 is the value of the gold in the coin

Explanation:

Diameter is 2r, the ratio of the coin is = 1.62cm / 2 = 0.81cm

Following the formula of the volume of the coin:

V = π r² h

V = π*(0.81cm)²*0.085cm

V = 0.172cm³ = mL

As the density of the gold is 19.32g/mL, the mass of 0.1752mL of gold is:

0.1752mL * (19.32g / mL) =

3.385g is the mass of the coin

As the price of the gold is $53.90 / g, the value of the gold in the coin is:

3.385g * ($53.90 / g) =

<h3>$182 is the value of the gold in the coin</h3>

7 0

3 years ago

Which of the following would force the reaction to favor the products? CaCO3(s) + 2H3O(aq) + Ca(aq) + 3H2O(1) + CO2(g) A) Removi

iVinArrow [24]

Answer: Option (C) is the correct answer.

Explanation:

As the given reaction is as follows.

$CaCO_{3}(s) + 2H_{3}O^{+}(aq) \rightleftharpoons Ca^{2+}(aq) + 3H_{2}O(l) + CO_{2}(g)$

According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.

Hence, in order to favor the formation of products when we remove $CO_{2}$ then there will occur decrease in it's concentration.

As $CO_{2}$ is forming on the product side and equilibrium will shift in the direction where there is less stress.

Hence, then equilibrium will shift in the forward direction, that is, on the products side.

Thus, we can conclude that removing $CO_{2}$ as it is formed would force the reaction to favor the products.

3 0

3 years ago