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Ganezh [65]
2 years ago
10

Two structures found in a plant cell but not in an animal cells are​

Chemistry
1 answer:
morpeh [17]2 years ago
6 0

Answer:

The plant cell has a cell wall, chloroplasts, plastids, and a central vacuole—structures not found in animal cells.

Explanation:

that is my answer

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You drop a rock weighing 23.2 g into a graduated cylinder that contains 55 mL. The level of the water rises to 62 mL. What is th
ra1l [238]

The density of the rock is 3.314g/mL

CALCULATE DENSITY:

  • According to this question, a rock weighs 23.2g. After dropping the rock into a graduated cylinder containing 55mL of water, the level changes to 62mL.

  • This means that the volume of the rock can be calculated as follows:

Volume of rock = 62mL - 55mL

Volume of rock = 7mL

Density can be calculated using the formula as follows:

Density = mass ÷ volume

Density = 23.2 ÷ 7

Density = 3.314g/mL

Therefore, the density of the rock is 3.314g/mL

Learn more: brainly.com/question/6034174?referrer=searchResults

4 0
2 years ago
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
2 years ago
A alkaline earth metal is an element is groups 3-12 of the periodic table true or false?
Karo-lina-s [1.5K]

False Alkaline earth metals is group 2

5 0
2 years ago
How do I do question number 4?
lisabon 2012 [21]
I hope you understood
hit me up if you have any other questions :>

6 0
2 years ago
What was the oxygen crisis
salantis [7]
The Great Oxidation Event (GOE), sometimes also called the Great Oxygenation Event, Oxygen Catastrophe, Oxygen Crisis, Oxygen Holocaust,[2] or Oxygen Revolution, was a time period when the Earth's atmosphere and the shallow ocean first experienced a rise in oxygen, approximately 2.4 billion years ago (2.4 Ga) to 2.1–2.0 Ga during the Paleoproterozoic era.[3] Geological, isotopic, and chemical evidence suggests that biologically produced molecular oxygen (dioxygen, O2) started to accumulate in Earth's atmosphere and changed Earth's atmosphere from a weakly reducing atmosphere to an oxidizing atmosphere,[4] causing many existing species on Earth to die out.[5] The cyanobacteria producing the oxygen caused the event which enabled the subsequent development of multicellular forms.
8 0
3 years ago
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