6. What is the number of moles of KF in a 29 gram sample?

The pH of an acidic solution is 2.47. what is the [H+]

solmaris [256]

$pH=-\log[H^{+}]\\\\\implies [H^{+}] =10^{-pH}=10^{-2.47}=0.0034~M$

8 0

3 years ago

How much heat is absorbed to change 13.8 grams of H2O from a solid to a liquid at zero degrees Celsius? ΔHfusion = 6.03 kJ/mol Δ

Delicious77 [7]

1 mole of H2O weighs 18 g

therefore 13.8 g of liquid H2O = 13.8/18 moles

ΔHvaporization = 40.65 kJ/mol

heat required to change 13.8 grams of H2O from a liquid to a gas at 100 degrees Celsius = 40.65 x 13.8/18 = 31.165 kJ

8 0

3 years ago

Read 2 more answers

The vapor pressure of water is 23.76 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in water is sucrose. Calculate

MrRissso [65]

Answer : The vapor pressure of solution is 23.67 mmHg.

Solution:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent (water) = 23.76 mmHg

$p_s$ = vapor pressure of solution= ?

$w_2$ = mass of solute (sucrose) = 12.25 g

$w_1$ = mass of solvent (water) = 176.3 g

$M_1$ = molar mass of solvent (water) = 18.02 g/mole

$M_2$ = molar mass of solute (sucrose) = 342.3 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{23.76-p_s}{23.76}=\frac{12.25\times 18.02}{176.3\times 342.3}$

$p_s=23.67mmHg$

Therefore, the vapor pressure of solution is 23.67 mmHg.

7 0

3 years ago

What is the most important scientific process

kotykmax [81]

Answer:

atoms bonding

Explanation:

6 0

4 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 $CO_2$

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

$\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g$

Theoretical yield of $CO_2$ = 0.572 g

Experimental yield of $CO_2$ = 0.391 g

Now we have to calculate the percent yield of $CO_2$

$\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100$

$\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%$

Therefore, the percent yield of $CO_2$ is, 68.4 %

6 0

3 years ago