All categories

2 years ago

Does anyone know the answer to this riddle???

Chemistry

2 answers:

BartSMP [9]2 years ago

6 0

Answer:

babe where are u???????!

Explanation:

babe plz text me im lowkey freaking out rn

ddd [48]2 years ago

6 0

Answer:

wind or helium or something balloons??

Explanation:

You might be interested in

Would you rather be hot or cold

Norma-Jean [14]

Answer:

cold

Explanation:

because being cold helps to refresh my body personally

4 0

2 years ago

Read 2 more answers

When ammonium chloride crystals are dissolved in water, the temperature of the water decreases. What does the temperature change

sertanlavr [38]

Answer:

See Explanation

Explanation:

A decreasing temperature indicates that the dissolution process for the ammonium chloride requires input of energy from surroundings. That is, the process is essentially 2 parts => system (object of interest - NH₄Cl) and the surroundings (everything else - solvent - H₂O). The surroundings (water) solvent is showing a <u>measured</u> decrease in temperature or loss of energy (exothermic to surroundings) which flows into the system (NH₄Cl) and effects dissolution of salt into solution (endothermic to system).

3 0

3 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 $NaCl$