Two types of stems are herbaceous and flexible true or false

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