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klemol [59]
4 years ago
15

Two types of stems are herbaceous and flexible true or false

Chemistry
1 answer:
Sav [38]4 years ago
8 0
False.

The two types of stems are herbaceous and WOODY stems.

Hope this helps! :D
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How many valence electrons are represented in the following electron configuration? 1s²2s²2p63s²3p4 2 4 6​
Svetlanka [38]

Answer:

we add only the the same highest enegy levels :3s&3p=2+4=6

7 0
2 years ago
When a tip of a plant is cut it doesn't grow well​
aleksandr82 [10.1K]

Explanation:

the tip of a growing plant contains special rapidly diving cells called apical meristem ,these cells are responsible for increase in the length of the plant . if we cut out these cells ,length growth of the plant will be stunned as these cells are not present anyplace else.

8 0
3 years ago
Read 2 more answers
How to find orbital notation?
AveGali [126]

Orbital notation is a way of writing an electron configuration to provide more specific information about the electrons in an atom of an element.

Orbital notation can be used to determine the quantum numbers of an electron.


6 0
3 years ago
Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C3H8, is burned with 75.0 g of oxygen.
kumpel [21]

Answer : The limiting reactant is O_2

Explanation : Given,

Mass of C_3H_8 = 30.0 g

Mass of O_2 = 75.0 g

Molar mass of C_3H_8 = 44 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_3H_8 and O_2.

\text{ Moles of }C_3H_8=\frac{\text{ Mass of }C_3H_8}{\text{ Molar mass of }C_3H_8}=\frac{30.0g}{44g/mole}=0.682moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{75.0g}{32g/mole}=2.34moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From the balanced reaction we conclude that

As, 5 mole of O_2 react with 1 mole of C_3H_8

So, 2.34 moles of O_2 react with \frac{2.34}{5}\times 1=0.468 moles of C_3H_8

From this we conclude that, C_3H_8 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Therefore, the limiting reactant is O_2

5 0
3 years ago
Convert 6.35 grams of aluminum sulfate to moles​
vampirchik [111]

Answer:

There are 0.0186 moles of formula units in 6.35 grams of aluminum sulfate \rm Al_2(SO_4)_3.

Explanation:

What's the empirical formula of aluminum sulfate?

Sulfate is an anion with a charge of -2 per ion. When sulfate ions are bonded to metals, the compound is likely ionic.

Aluminum is a group III metal. Its ions tend to carry a charge of +3 per ion.

The empirical formula of an ionic compound shall balance the charge on ions with as few ions as possible.

The least common multiple of 2 and 3 is 6. That is:

  • Three sulfate ions \rm {SO_4}^{2-} will give a charge of -6.
  • Two aluminum ions \rm Al^{3+} will give a charge of +6.

Pairing three \rm {SO_4}^{2-} ions with two \rm Al^{3+} will balance the charge. Hence the empirical formula: \rm Al_2(SO_4)_3.

What's the mass of one mole of aluminum sulfate? In other words, what's the formula mass of \rm Al_2(SO_4)_3?

Refer to a modern periodic table for relative atomic mass data:

  • Al: 26.982;
  • S: 32.06;
  • O: 15.999.

There are

  • two Al,
  • three S, and
  • twelve O

in one formula unit of \rm Al_2(SO_4)_3.

Hence the formula mass of \rm Al_2(SO_4)_3:

\underbrace{2\times 26.982}_{\rm Al} + \underbrace{3\times 32.06}_{\rm S} + \underbrace{12\times 15.999}_{\rm O} = \rm 342.132\;g\cdot mol^{-1}.

How many moles of formula units in 6.35 grams of \rm Al_2(SO_4)_3?

\displaystyle n = \frac{m}{M} = \rm \frac{6.35\;g}{342.132\;g\cdot mol^{-1}} = 0.0186\;mol.

8 0
4 years ago
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