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Nina [5.8K]
3 years ago
14

If you wanted to look at a cell wall.Which type of cell wall would you observe?

Chemistry
1 answer:
nirvana33 [79]3 years ago
3 0

If you are trying to observe a cell wall, you should look at a plant cell. There are only two different types of cells, which are plant and animal cells. The plant cell is the only one which contains a cell wall. Animal cells actually only have a <em>cell membrane. </em>

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kolbaska11 [484]
Aluminum or glass I think
3 0
2 years ago
Read 2 more answers
Is condensation an exothermic or endothermic process? Based on Le Cha^telier’s principle, would you predict the vapor pressure o
Irina18 [472]

Explanation:

A process in which water vapor changes into liquid state is known as condensation. As we know that when energy is released in a reaction then it is known as exothermic reaction and when energy is absorbed in a reaction then it is known as endothermic reaction.

As vapors have high energy so, when they change into liquid state then heat energy is released by them. Therefore, condensation is an exothermic reaction.

As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.

So, when there will occur a decrease in temperature then molecules of a gas will come closer to each other. Hence, there will also occur a decrease in vapor pressure of the gas.

7 0
3 years ago
How many helium nuclei fuse together when making carbon?
Gemiola [76]
The answer is the 3. Hope this helped!
8 0
3 years ago
In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

6 0
3 years ago
How many moles of nitrogen trifluoride (NF3) can be produced from 9.65 mole of Fluorine gas (F2)
user100 [1]

Answer:

6.43 moles of NF₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.

Thus, 6.43 moles of NF₃ were obtained from the reaction.

4 0
3 years ago
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