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2 years ago

Refer to the image please!

Chemistry

1 answer:

Ludmilka [50]2 years ago

8 0

Answer:

9. Electron Affinity

10. The second option....

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Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/

Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of $C_2H_4$ and $H_2$ = 25.0 atm

Temperature of $C_2H_4$ and $H_2$ = $250^oC=273+250=523K$

Volume of $C_2H_4$ = 1050 L per min

Volume of $H_2$ = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of $C_2H_6$ = 30 g/mole

First we have to calculate the moles of $C_2H_4$ and $H_2$ by using ideal gas equation.

For $C_2H_4$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=611.34moles$

For $H_2$ :

$PV=nRT\\\\n=\frac{PV}{RT}$

$n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}$

$n=902.46moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_2H_4+H_2\rightarrow C_2H_6$

From the balanced reaction we conclude that

As, 1 mole of $C_2H_4$ react with 1 mole of $H_2$

So, 611.34 mole of $C_2H_4$ react with 611.34 mole of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $C_2H_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $C_2H_6$ .

As, 1 mole of $C_2H_4$ react to give 1 mole of $C_2H_6$

As, 611.34 mole of $C_2H_4$ react to give 611.34 mole of $C_2H_6$

Now we have to calculate the mass of $C_2H_6$ .

$\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6$

$\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g$

The theoretical yield of $C_2H_6$ = 18340.2 g

The actual yield of $C_2H_6$ = 14.0 kg = 14000 g (1 kg = 1000 g)

Now we have to calculate the percent yield of $C_2H_6$

$\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%$

Therefore, the percent yield of the reaction is, 76.34 %

5 0

3 years ago

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

alekssr [168]

Answer:

208.7°C was the initial temperature of the limestone.

Explanation:

Heat lost by limestone will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of limestone = $m_1=62.6 g$

Specific heat capacity of limestone = $c_1=0.921 J/g^oC$

Initial temperature of the limestone = $T_1=?$

Final temperature = $T_2$ =T = 51.9°C

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=23.1^oC$

Final temperature of water = $T_2$ =T = 51.9°C

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(62.6 g\times 0.921 J/g^oC\times (51.9^oC-T_1))=75.0 g\times 4.186 J/g^oC\times (51.9^oC-23.1^oC)$

$T_1=208.7^oC$

208.7°C was the initial temperature of the limestone.

7 0

3 years ago

What two colors make pink?

KiRa [710]

Red and white I think

5 0

3 years ago

Read 2 more answers

The flask contains 10.0 mL of HCl and a few drops of phenolphthalein indicator. The buret contains 0.160 M NaOH. It requires 18.

olchik [2.2K]

Answer:

Approximately $0.291\; \rm M$ (rounded to two significant figures.)

Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

5 0

3 years ago

How many moles of co2 are produced when 5.70 mol of ethane are burned in an excess of oxygen

sweet-ann [11.9K]

I believe the answer is 6

4 0

3 years ago