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lorasvet [3.4K]
3 years ago
15

From where do the placenta and umbilical cord develop?

Chemistry
2 answers:
Anastasy [175]3 years ago
8 0
It’s either the first or second one

I think it’s the first one - the outer cells of the blastocyst
frozen [14]3 years ago
5 0

Answer:

When an egg is fertilized, it divides into two components — one becomes the embryo, which develops into the fetus, and the other becomes the placenta, which grows along the lining of the uterus. The umbilical cord develops from embryonic tissue and will grow about 60 centimetres long.

The umbilical cord develops from and contains remnants of the yolk sac and allantois. It forms by the fifth week of development, replacing the yolk sac as the source of nutrients for the embryo.

OR put simply

A, the oyster cells of the blastocyst.

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Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KClO3(s)2KCl(s) + 3O2(g) The prod
Ipatiy [6.2K]

Answer:

The number of moles of KClO₃ reacted was 0,15 mol

Explanation:

For the reaction:

2KClO₃(s) → 2KCl(s) + 2O₂(g)

The only gas product is O₂.

Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:

749mmHg - 23,8mmHg = 725,2mmHg

Using gas law:

PV/RT = n

Where:

P is pressure (725,2mmHg ≡ <em>0,9542atm</em>)

V is volume (<em>5,76L</em>)

R is gas constant (<em>0,082 atmL/molK</em>)

And T is temperature (25°C ≡ <em>298,15K</em>)

Replacing, number of moles of O₂ are <em>0,2248 moles</em>

As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:

0,2248 mol O₂×\frac{2 mol KClO_{3}}{3 mol O_{2}} = <em>0,15 mol of KClO₃</em>

I hope it helps!

8 0
3 years ago
A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus
SIZIF [17.4K]
Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal

<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>
8 0
3 years ago
At what temperature is the following reaction feasible: HCl(g) + NH3(g) -&gt; NH4Cl(s)?
Nutka1998 [239]
Energy is distributed not just in translational KE, but also in rotation, vibration and also distributed in electronic energy levels (if input great enough, bond breaks).

All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.

Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.

Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.

We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:

ΔSθtot = ΔSθsys +  ΔSθsurr

ΔSθtot must be >=0 for a chemical change to be feasible.

For example: CaCO3(s) ==> CaO(s) + CO2(g) 

ΔSθsys = ΣSθproducts – ΣSθreactants 

ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) 

ΔSθsurr is –ΔHθ/T(K) and ΔH is very endothermic (very +ve),

Now ΔSθsys is approximately constant with temperature and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall.

But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800oCΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.

CaCO3(s) ==> CaO(s) + CO2(g)  ΔHθ = +179 kJ mol–1  (very endothermic)

This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1
ΔSθsurr is –ΔHθ/T = –(179000/T)
ΔSθtot = ΔSθsys +  ΔSθsurr
ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T
If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.
For T = 298K (room temperature)

ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change

For T = 500K (fairly high temperature for an industrial process)

ΔSθtot = 161 – 179000/500 = –197.0, still no good

For T = 1200K (limekiln temperature)

ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change

Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.
ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K

This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.

8 0
2 years ago
Read 2 more answers
Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to th
Luba_88 [7]

Answer:

The percent yield of the reaction is 82%

Explanation:

First step: make the chemist equation.

2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)

As the statement says that aluminun is in excess, the limiting reactant is the Fe2O3

Second step: Find out the moles in the reactant.

Molar weight Fe2O3: 159.7 g/m

Mass / Molar weight = moles

50 g /159.7 g/m = 0.313 moles

Third step: Analyse the reaction. 1 mol of Fe2O3 makes 2 moles of Fe.

1 mol Fe2O3 ____ 2Fe

0.313 mol Fe2O3 ____ 0.626 moles

Molar weight Fe = 55.85 g/m

Moles . molar weight = mass

55.85g/m . 0.626m = 34.9 grams

This will be the 100% yield of the reaction but we only made 28.65 g

34.9 g ____ 100%

28.65 g ____ 82.09 %

3 0
3 years ago
Read 2 more answers
How to calculate the percentage of a anhydrous salt
Anna35 [415]
Yep u right bro but like you know I’m here with these guys you know what I mean tho
8 0
3 years ago
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