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Leviafan [203]
2 years ago
10

How can i make babies

Chemistry
2 answers:
Studentka2010 [4]2 years ago
8 0

Answer with Explanation:

Babies are formed when the sperm cell from the male testicular gonad meets with the women's egg cell (from the ovaries). This penetration causes the<em> "fertilization process." </em>This naturally occurs two weeks from the woman's last day of menstruation. The fertilization allows the egg to rapidly divide. This is followed by "implantation," a process whereby the blastocyte attaches to the woman's endometrium. <u>Around three weeks, the blastocyte is formed into an embryo. </u>This will be called fetus once the woman reaches its 8th week of pregnancy.

This is a brief process regarding your question on how babies are formed.

vampirchik [111]2 years ago
7 0

Answer:

You will need an opposite gender to form a baby.

Explanation:

You might be interested in
How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a
Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of water = 72.0  g

c_{s} = specific heat of ice = 2.09J/g^0C

c_{l} = specific heat of liquid water = 4.184J/g^0C

n = number of moles of water = \frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles

\Delta H_{fusion} = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]\Delta H=34279.8J=34.3kJ        (1 KJ = 1000 J)

Therefore, the enthalpy change is 34.3 kJ

5 0
2 years ago
What should you do with a rag that has been used to wipe up spilled gasoline? a)Place it in the bilge. b)Hang it over the gunwal
Tcecarenko [31]

Answer:

A rag contaminated with gasoline is considered a hazard waste. The question seems to be asked to ask for indications of rag disposal on a ship, and if so, the answer is d) Discard it on land BUT with special precautions

Explanation:

Gasoline is a highly flammable organic solvent that is used as fuel. For the above is a dangerous substance. while the ship reaches the mainland, the rag must be stored avoiding contact with environmental agents such as the sun, pets or food. Once in a dry land the rag can be delivery for accurate disposal.

I hope my answer helps you

6 0
2 years ago
Write the name for: Mg3O2
sveticcg [70]

Answer:

you sure it is the right sign your using

4 0
2 years ago
This question deals with waste disposal in the Solutions and Spectroscopy experiment. What should be done to waste solutions con
Korvikt [17]

Answer:

b. It should be dumped in a beaker labeled "waste copper" on one's bench during the experiment.

d. It should be disposed of in the bottle for waste copper ion when work is completed.

Explanation:

Solutions containing copper ion should never be disposed of by dumping them in a sink or in common trash cans, because this will cause pollution in rivers, lakes and seas, being a contaminating agent to both human beings and animals. They should be placed in appropriate compatible containers that can be hermetically sealed. The sealed containers must be labeled with the name and class of hazardous substance they contain and the date they were generated.

It never should be returned to the bottle containing the solution, since it can contaminate the solution of the bottle.

In the Solutions and Spectroscopy experiments there is always wastes.

3 0
3 years ago
How long does it take to electroplate 0.5 mm of gold on an object with a surface area of 31 cm^^ from an Au3+(aq) solution with
konstantin123 [22]

Answer:

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

Explanation:

Mass of gold = m

Volume of gold = v

Surface area on which gold is plated = a=31 cm^2

Thickness of the gold plating  = h = 0.5 mm = 0.05 cm

1 mm = 0.1 cm

V=a\times h=31 cm^2\times 0.05 cm=1.55 cm^3

Density of the gold = d=19.3 g/cm^3

m=d\times v=19.3 g/cm^3\times 1.55 cm^3=29.915g

Moles of gold = \frac{29.915 g}{197 g/mol}=0.152 mol

Au^{3+}+3e^-\rightarrow Au

According to reaction, 1 mole of gold required 3 moles of electrons,then 0.152 moles of gold will require :

\frac{3}{1}\times 0.152 mol=0.456 mol of electrons

Number of electrons = N =0.456\times \times 6.022\times 10^{23}

Charge on single electron = q=1.6\times 10^{-19} C

Total charge required = Q

Q=N\times q

Amount of current passes = I = 8 Ampere

Duration of time  = T

I=\frac{Q}{T}

T=\frac{N\times q}{I}

=\frac{0.456\times \times 6.022\times 10^{23}\times 1.6\times 10^{-19} C}{8 A}=5492 s

It will take 5492 seconds to electroplate 0.5 mm of gold on an object .

7 0
2 years ago
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