Please help asap!!!!!!!!!!! How many molecules are present in 0.5 moles of NaOH?

qaws [65]

D-1 electron in their outer energy level.

6 0

2 years ago

according to the collision theory, for a reaction to be successful, it must have correct orientation of the atoms of the reactan

DerKrebs [107]

The collision must have sufficient energy as well. If sufficient energy is not present the activation energy will not be over come and the bonds in the reactants will not be broken.

I hope this helps. Let me know if anything is unclear.

6 0

3 years ago

Read 2 more answers

Chem Muti Choice. Tell me the correct answer.

Alex787 [66]

Answer:

my gues is red not a 100% but its in the 700s

Explanation:

3 0

3 years ago

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Describe the benefits and the drawback of natural gas.

Finger [1]

When looking to power your equipment or vehicle with natural gas, the first question that springs to mind is: what is natural gas? A lot of people use natural gas in their homes for cooking and heating, but they don’t really give it some thought. So, let’s see what natural gas is and how it’s different from other forms of fossil fuels like oil and coal.

Natural Gas is a fossil fuel that exist in a gaseous state and is composed mainly of methane (CH4) a small percentage of other hydrocarbons (e.g. ethane). The use of natural gas is becoming more and more popular as it can be used with commercial, industrial, electric power generation and residential applications.

5 0

2 years ago

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 $NaCl$