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3 years ago

How many moles of O2 are needed to produce 50.0 moles of CO2?

Chemistry

1 answer:

-Dominant- [34]3 years ago

6 0

Answer:

4 moles

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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if an electron is released during radioactive decay which type of Decay has taken place a gamma decay b beta decay c electromagn

olasank [31]

Answer:

i hope you have a great day!

Explanation:

3 0

3 years ago

Read 2 more answers

If gas in a sealed container has a pressure of 50 kpa at 300 k, what will the pressure be if the temperature rises to 360 k? 161

Illusion [34]

$60 \; \text{kPa}$

The pressure an ideal gas exerts on a sealed container of fixed volume is directly related to its temperature in degrees Kelvin. That is:

$T_1 / T_2 = P_1 / P_2$

In this scenario:

$T_1 = 300 \; \text{K}$
$P_1 = 50 \; \text{kPa}$
$T_2 = 360 \; \text{K}$

Rearranging gives

$\begin{array}{lll}P_2 &=& P_1 \cdot T_2 / T_1 \\ &=& 50 \; \text{kPa} \times (360 \; \text{K})/ (300 \; \text{K})\\ &=& 60 \; \text{kPa}\end{array}$

6 0

2 years ago

what is the molarity of an HCI solution if 25.0 ml of 0.185 M NaOH is required to neutralize 0.0200 L of HCI?

butalik [34]

Answer:

Molarity of HCl solution = 0.25 M

Explanation:

Given data:

Volume of NaOH= V₁ = 25.0 mL (25/1000 = 0.025 L)

Molarity of NaOH solution=M₁ = 0.185 M

Volume of HCl solution = V₂ = 0.0200 L

Molarity of HCl solution = M₂= ?

Solution:

M₁V₁ = M₂V₂

0.185 M ×0.025 L = M₂ × 0.0200 L

M₂ = 0.185 M ×0.025 L / 0.0200 L

M₂ = 0.005M.L /0.0200 L

M₂ = 0.25 M

7 0

3 years ago

Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating

kvv77 [185]

<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

$1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol$

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

$\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }$

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = $\frac{1}{1}\times 0.0295=0.0295moles$ of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g$

To calculate the percentage yield of iodo-salicylamide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

$\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%$

Hence, the percent yield of the reaction is 68.68%.

4 0

3 years ago

HELLLPPPP!!!!

AleksAgata [21]

A. Because you have to simplify

6 0

3 years ago