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guapka [62]
3 years ago
7

What is the first step of a ladybug during the growth and development process​

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

Damian here! (ノ◕ヮ◕)ノ*:・゚✧

The newly hatched larva is in its first instar, a developmental stage that occurs between molts. It feeds until it grows too big for its cuticle, or soft shell, and then it molts. After molting, the larva is in the second instar. Ladybug larvae usually molt through four instars, or larval stages, before preparing to pupate.

Explanation:

hope this helps? :))

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A flask containing helium gas is connected to an open-ended mercury manometer. The open end is exposed to the atmosphere, where
stepladder [879]

Answer:

726 torr

Explanation:

Generally, atmospheric pressure can be measured using a manometer which is in form of a U-shaped tube. In addition, 1 mm Hg is equivalent to 1 torr. Therefore, 752 torr is equivalent to 752 mm Hg. Therefore, the total pressure will be equivalent to the atmospheric pressure (mm Hg) + the mercury height.

In this case, the mercury height = -26 mm

Thus:

The helium pressure = 752 - 26 = 726 mm Hg

This is also equivalent to 726 torr

8 0
3 years ago
Calculate the density of chloroform
Rina8888 [55]

Answer: The density of chloroform is 1.47 g/mL

Explanation : Given,

Volume = 40.5 mL

Mass of cylinder = 85.16 g

Mass of cylinder and liquid = 145.10 g

First we have to calculate the mass of liquid (chloroform).

Mass of liquid = Mass of cylinder and liquid - Mass of cylinder

Mass of liquid = 145.10 g - 85.6 g

Mass of liquid = 59.5 g

Now we have to calculate the density of liquid (chloroform).

Formula used:

Density=\frac{Mass}{Volume}

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

Density=\frac{59.5g}{40.5mL}

Density=1.47g/mL

Therefore, the density of chloroform is 1.47 g/mL

5 0
2 years ago
Calculate the heat energy released when 21.1 g of liquid mercury at 25.00 °C is converted to solid mercury at its melting point.
PSYCHO15rus [73]
Heat capacity yea urjjrjrjrbdnsnnend
3 0
3 years ago
What is the chemical composition of the mobile phase in this experiment?
Alecsey [184]
H2SO.Mgslfurmobile phase in this experiment
6 0
2 years ago
The heat of reaction for the combustion of propane is –2,045 kJ. This reaction is: C3H8(g) +502 (g) 3CO2 (g) + 4H2O (g). Determi
Ede4ka [16]

Answer:

\Delta _fH_{C_3H_8}=-102.7kJ/mol

Explanation:

Hello,

In this case, we can consider that the given heat of combustion is indeed the heat of reaction since it corresponds to the combustion of propane, which is computed by using the heat formation of all the involved species as shown below:

\Delta _cH=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-5*\Delta _fH_{O_2}

Thus, since the heat of formation of gaseous carbon dioxide is -393.5 kJ/mol, water -241.8 kJ/mol and oxygen 0 kJ/mol, the heat of formation of propane is:

\Delta _fH_{C_3H_8}=3*\Delta _fH_{CO_2}+4*\Delta _fH_{H_2O}-5*\Delta _fH_{O_2}-\Delta _cH\\\\\Delta _fH_{C_3H_8}=3*(-393.5)+4*(-241.8)-5*0-(-2045)\\\\\Delta _fH_{C_3H_8}=-102.7kJ/mol

Best regards.

8 0
3 years ago
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