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andriy [413]
2 years ago
9

a student takes two strips from the epidermis of an onion and places one in distilled water and the other in concreted salt solu

tion. She then uses the camera to photograph a sample of these cells under a microscope. Suggest why a red onion is often used when carrying out this investigation
Chemistry
1 answer:
maks197457 [2]2 years ago
5 0

see bahubali picture and pushpa this is your answer Visit your school library, read newspapers and discuss with your elders in your family to get information related to various activities, campaigns and news items about animal welfare. You can also do web search using keywords like endangered animals, animal rights, animal welfare, etc. Write a note of about 300-400 words describing the facts on the animal. Draw/paste picture as suitable. Note the following points: 1. animal in concern 2. year of origin of the project/activities/campaign, etc. 3. information regarding the animal 4. human-animal relationship

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Question 1 (1 point)<br> ROOTS--What is NOT a root function in plants?
Jlenok [28]

Answer:

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Explanation:

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3 0
3 years ago
What is the total number of atoms in a 75 g sample of zinc?
velikii [3]
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8 0
2 years ago
Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of
harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g)   (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

n = \frac{m}{M}

Where:    

m: is the mass

M: is the molar mass

  • For <em>hydrogen </em>we have:

n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles

  • And for <em>nitrogen</em>:

n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles

We can see in reaction (1) that <u>3 moles of hydrogen</u> react with <u>1 mol of nitrogen</u>, so the number of hydrogen moles needed to react nitrogen is:

n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles

Since we have <u>3.47 moles of hydrogen</u> and we need <u>7.50 moles</u> to react with all the mass of nitrogen, the <em>limiting reactant</em> is <em>hydrogen</em>.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that <u>3 moles of hydrogen</u> produces <u>2 moles of ammonia</u>, so:

n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles

Hence, hydrogen would produce <u>2.31 moles of ammonia</u>.

Therefore, hydrogen is the limiting reactant because <u>7.5 moles of hydrogen would be needed to consume the available nitrogen</u> (option 1).

Find more about limiting reactants here:

brainly.com/question/2948214?referrer=searchResults

   

I hope it helps you!                        

6 0
2 years ago
Draw the structure for 1-nitrobutane, making sure to add all non-zero formal charges.
TiliK225 [7]

The molecular structure of 1-nitrobutane is C_{4} H_{9} NO_{2}. The structure of 1-nitrobutane is shown below.

An atom's formal charge would be determined by the covalent model of chemical bonding, which assumes that almost all chemical bonds include equal sharing of electrons among all atoms, regardless their relative electronegativity.

The structure for 1-nitrobutane, making sure to add all non-zero formal charges

There are four kind of molecule present in 1-nitrobutane and they are carbon, hydrogen , nitrogen and oxygen. Nitrogen is bonded with two oxygen atom out of them one oxygen atom is attached with single bond and second oxygen atom is bonded with double bond. Nitrogen has positive charge whereas oxygen has negative charge.

It is a kind of alkane in with nitro group is attached with alkane group.

To know more about 1-nitrobutane

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7 0
1 year ago
if 45.0 ml of 1.50 M Ca(OH)2 are needed to neutralize 25.0 ml of HI of unknown concentration, what is the molarity of the HI?
ipn [44]

Answer:

M of HI = 5.4 M.

Explanation:

  • We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

<em>(XMV) acid = (XMV) base.</em>

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

<em>(XMV) HI = (XMV) Ca(OH)₂.</em>

For HI; X = 1, M = ??? M, V = 25.0 mL.

For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.

<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>

6 0
3 years ago
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