How many valence electrons does B (boron) have? A. 13 B. 5 C. 10 D. 3

Characteristics of the fungus kingdom (all of them)

Nana76 [90]

- Fungi are "eukaryotic heterotrophs"

- Cell Walls contain a substance called Chitin

- All fungi are multicellular except yeast.

- Fungi are made up of Hyphae

- One hypha is only 1 cell thick

- The most common fungi is the DECOMPOSER

5 0

3 years ago

How many grams of Zn are in 0.356 mole of Zn?

Darina [25.2K]

Answer:

23.3 g

Explanation:

0.356 mole × 65 = 23.3 g

6 0

3 years ago

What element from the "Periodic Table" is named after an "Austrian Scientist"??

soldier1979 [14.2K]

Answer:

Should be either mendele or fermium

maybe lawrencium not sure but I think it’s lawrencium! Sorry if wrong heh

Explanation: really trying to help! If it’s wrong super sorry!!!

3 0

3 years ago

A can was filled with crushed ice, sealed, and massed. The ice was melted by slowly warming the can and its contents. No water v

Lera25 [3.4K]

Answer:

A) The mass would be the same.

Explanation:

Since there is no loss of any particle to vapor during the phase change process from solid to liquid, the mass of the before and after the process will remain the same.

In this way, the law of conservation of mass is obeyed.
Mass is the amount of matter contained in a substance.
Since there is no room for escape or matter loss, the mass will remain the same.

8 0

3 years ago

Read 2 more answers

A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi

Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

Mass of $Na$ = 5.0 g

Mass of $Cl_2$ = 10.0 g

Molar mass of $Na$ = 23 g/mol

Molar mass of $Cl_2$ = 71 g/mol

First we have to calculate the moles of $Na$ and $Cl_2$ .

$\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}$

$\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{10.0g}{71g/mol}=0.141mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

$2Na+Cl_2\rightarrow 2NaCl$

From the balanced reaction we conclude that

As, 2 mole of $Na$ react with 1 mole of $Cl_2$

So, 0.217 moles of $Na$ react with $\frac{0.217}{2}=0.108$ moles of $Cl_2$

From this we conclude that, $Cl_2$ is an excess reagent because the given moles are greater than the required moles and $Na$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$