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4 years ago

Which of the following cannot pass across the placenta?

Chemistry

2 answers:

zmey [24]4 years ago

7 0

Oxygen cannot pass through the placenta

S_A_V [24]4 years ago

3 0

The placenta is the structure that allows for transfer of wastes, oxygen, and
nutrients between the mother and the fetus. The fetal blood comes extremely
close to the maternal blood in the placenta, but <em>there is no intermingling of fetal
and maternal <u>blood</u></em>. The fetus and mother can even have, and often do have,
different blood types.

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What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1? 6.4 × 10−4 g 6.4 × 10−3 g

eduard

Compete Question:

What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?

Passage: "16 mmol of CDP in 1 L of buffer"

Answer:

6.4 × 10-2 g

Explanation:

$Mass = Mole × Molar Mass$

we are given from the question that 16 mmol of CDP is in 1 L of buffer

this mean that we have $16 × 10^-3$ moles of CDP in 1 liter of buffer.

so the mass of CDP in one liter of buffer will be calculate as,

mass of CDP = $16 × 10^-3$ × 403g mol−1

= $64 × 10^-1$

= 6.4 g/L

But because the question

asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:

6.4 g/L × 10 mL

6.4 g/L × 0.01 L = $6.4 × 10^-2$

8 0

4 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

4 years ago

A watt is a unit of power or energy per unit time (1 W = 1 J/s). A semiconductor laser in a CD player has an output wavelength o

GREYUIT [131]

Answer:

n = 2.208x10¹⁸ photons

Explanation:

The energy of a photon( an elementary particle) is given by the equation:

E = nxhxf

Where n is the number of photons, h is plank constant (6,62x10⁻³⁴ J.s), and f is the frequency. Knowing that the power level is 0.120mW (1.2x10⁻⁴ W), the energy in J, for a time of 78 min (4680 s)

E = 1.2x10⁻⁴x4680 = 0.5616 J

The frequency of a photon is its velocity ( c= 3x10⁸ m/s) divided by its wavelength, which is 780 nm = 780x10⁻⁹ m

f = 3x10⁸/780x10⁻⁹

f = 3.846x10¹⁴ s⁻¹

Then, the number of photons is:

0.5616 = nx6,62x10⁻³⁴x3.846x10¹⁴

n = 2.208x10¹⁸ photons.

5 0

3 years ago

Find the cell potential for a system whose ∆G = +55 kJ and 3 moles of electrons are exchanged.

rosijanka [135]

Answer:- cell potential = -0.19 volts

Solution:- The equation that shows the connection between $\Delta G$ and cell potential, E is written as:

$\Delta G=-nFE$

in this equation, n stands for moles of electrons, E stands for cell potential and F stands for faraday constant and it's value is $\frac{96485C}{mol}$ .