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4 years ago

Explain what is meant by Pleistocene ice age

1 answer:

7 0

By Pleistocene ice age is meant the glacial periods that appeared in the Pleistocene. The Pleistocene is the first epoch of the Quaternary. It is an epoch when there was a so called ''ice age'' on the Earth, or rather a glacial period. During this ice age the planet had much lower temperatures on a global scale. The climate was also much drier. Lot of ocean water was frozen in the ice sheets that were stretching deep into the North American and Eurasian continents, which resulted in much lower sea levels that today as well. The places further north than 40 degrees of latitude were almost exclusively covered with ice, so life was almost impossible apart from some coastline places.

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A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so

Oksana_A [137]

Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

Volume of solution = 100.0 mL

Molarity = 2.41 M

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}$

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

$2.41M=\frac{9.56g\times 1000}{\text{Molar mass of unknown substance}\times 100.0mL}$

$\text{Molar mass of unknown substance}=39.7g/mol$

Therefore, the molar mass of unknown substance is, 39.7 g/mol

5 0

3 years ago

Orbital structure of sodium?

algol13

Answer:

na.

sodium

11( 2,81,)

pls Mark me as brainlest

5 0

3 years ago

Aight imma let you guys i

katrin [286]

Answer:

Explanation:

it is right

6 0

3 years ago

Read 2 more answers

Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy

gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure , Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$ ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

$Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

$2P=PE^{6.4 \times 10^{-4}P}$

$2=E^{6.4 \times 10^{-4}P}$

$\ln 2 = 6.4 \times 10^{-4}P$

∴ P = 1083.04 atm

c). dG = Vdp -S dt at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp$

$=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

$=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol

7 0

3 years ago

When the volume of a gas is

neonofarm [45]

Charles law states that there is a directly proportional relationship between the volume and the temperature of the gas at certain pressure.

Therefore,

V1/T1 = V2/T2

V1 is unknown

T1 = 159K

V2 = 15.5m3

T2 = 456K

V1/159 = 15.5/456

V1 = (15.5*159)/456 = 5.404m3.

When the volume of a gas is changed from 5.404 m3 to 15.5 m3 the temperature will change from 159 K to 456 K.

5 0

3 years ago