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1 year ago

The life cycle of silkworm includes ------------- stage after the larvae stage

Chemistry

1 answer:

Ray Of Light [21]1 year ago

8 0

Answer:

Pupa Stage

Explanation:

Silk worm consists of four stages- the adult, the egg, the larva (caterpillar) and the pupa stage.

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Which of the following changes

Alisiya [41]

Exothermic reactions

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2 years ago

Consider the SCl2 molecule. (a) What is the electron config- uration of an isolated S atom? (b) What is the electron con- figura

iragen [17]

Answer:

(a) 1s² 2s² 2p⁶ 3s² 3p⁴

(b) 1s² 2s² 2p⁶ 3s² 3p⁵

(d) No valence orbital remains unhybridized.

Explanation:

<em>Consider the SCl₂ molecule. </em>

<em>(a) What is the electron configuration of an isolated S atom? </em>

S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.

<em>(b) What is the electron configuration of an isolated Cl atom? </em>

Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.

<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>

SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.

<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>

No valence orbital remains unhybridized.

7 0

2 years ago

The atmosphere is the air surrounding Earth and is made up of 78% _____, 21% oxygen, and other trace gases. Please help this is

oksano4ka [1.4K]

The Earth is made up of 78% of water!!

4 0

1 year ago

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

2 years ago

An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated

pashok25 [27]

Answer:

C₄H₂N₂

Explanation:

First we<u> calculate the moles of the gas</u>, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K

n = 0.0256 mol

Now we<u> calculate the molar mass of the compound</u>:

2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to<em> </em><u>calculate the empirical formula</u>:

C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4

H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2

N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2

So the empirical formula is C₄H₂N₂

6 0

2 years ago

The life cycle of silkworm includes ------------- stage after the larvae stage​

The life cycle of silkworm includes ------------- stage after the larvae stage