What is the mass of 9.67x10^23 molecules of water?

The oxidation number of all the element in free state is 0

Art [367]

Answer:

Zero-Nine

Explanation:

this is becasue these numbers are rather small and if you plug these numbers into an equation you will most likely get 0.

8 0

3 years ago

daniel has a sample of pure copper. its mass is 89.6 grams (g), and its volume is 10 cubic centimeters (cm3). what’s the density

77julia77 [94]

D = mass / volume

d = 89.6 / 10

d = 8.96 g/cm³

3 0

3 years ago

From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O

Drupady [299]

Answer: 125 g

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles$

The balanced reaction is:

$B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O$

According to stoichiometry :

1 mole of $B_2H_6$ require = 3 moles of $O_2$

Thus 1.30 moles of $B_2H_6$ will require= $\frac{3}{1}\times 1.30=3.90moles$ of $O_2$

Mass of $O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g$

Thus 125 g of $O_2$ will be needed to burn 36.1 g of $B_2H_6$

4 0

3 years ago

The initial volume of HCl was 1.25 ml and LiOH was 2.65 ml. The final volume of HCL was 13.60 ml and LiOH was 11.20 ml. If the L

Dennis_Churaev [7]

Q1)
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted - $\frac{0.140 mol/L*8.55mL}{1000mL}$ = 0.001197 mol
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of HCl moles in 12.35 mL - 0.001197 mol
Then number of HCl moles in 1000 mL - $\frac{0.001197*1000mL}{12.35mL}$
Molarity of HCl - 0.0969 M

Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL - $\frac{0.140mol*12.35 mL}{1000mL}$
Number of HCl moles reacted - 0.001729 mol
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL - $\frac{0.001729mol*1000mL}{8.55 mL}$
molarity of LiOH - 0.202 M

5 0

3 years ago

Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3

AveGali [126]

Answer : The percent yield of $CO_2$ is, 68.4 %

Solution : Given,

Mass of $CH_4$ = 0.16 g

Mass of $O_2$ = 0.84 g

Molar mass of $CH_4$ = 16 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $CH_4$ and $O_2$ .

$\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 0.026 moles of $O_2$ react with $\frac{0.026}{2}=0.013$ moles of $CH_4$

From this we conclude that, $CH_4$ 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.

Now we have to calculate the moles of $CO_2$

From the reaction, we conclude that

As, 2 mole of $O_2$ react to give 1 mole of $CO_2$

So, 0.026 moles of $O_2$ react to give $\frac{0.026}{2}=0.013$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$

$\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2$