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2 years ago

Can someone help me with these??

Chemistry

1 answer:

ryzh [129]2 years ago

3 0

Answer: let me see I gotta do the work first then I’ll give you the answers

Explanation:

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Combustion of hydrocarbons such as undecane (C_11H_24) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Eart

bekas [8.4K]

Answer:

Volume of carbon dioxide is 428.23 L.

Explanation:

Below is the chemical reaction or chemical equation for the combustion of hydrocarbons such as undecane into carbon dioxide.

$C_{11}H_{24}(l) + 17O_{2}(g) \to 11CO_{2} (g) + 12H_2O(g)$

Here, undecane is in liquid form that reacts with gaseous oxygen (combustion) and produces carbon dioxide and water as a product in the gaseous form.

The molar mass of undecane = $156.31 g/mol.$

$\text{Number of moles} = \frac{260g}{156.31g/mol} = 1.66 \ mol.$

From the equation, it can be seen that 1 mole of undecane produces 11 moles of carbon dioxide. Therefore, 1.66 mol will produce 18.26 mol of carbon dioxide.

Now find the volume of 18.26 mol of carbon dioxide when the temperature is 13 degrees Celsius and pressure is 1 atm.

$V = \frac{nRT}{P} \\$

$V = \frac{18.26 \times 0.082 \times 286 \ K}{1atm} \\$

$V = 428.23 \ L$

6 0

3 years ago

Near the equator, the sun heats the surface strongly, causing warm air to rise steadily. This creates the _________________, an

kenny6666 [7]

Doldrums, low is the correct answer.

8 0

2 years ago

Large-scale environmental catastrophes _______.

Ymorist [56]

Answer:

D. All of the Above

Explanation:

i just took the test on edgenuity

3 0

3 years ago

Determine the mass of CaCO3 required to produce 40.0 mL CO2 at STP. Hint use molar volume of an ideal gas (22.4 L)

cupoosta [38]

Answer:

$m_{CaCO_3}=0.179gCaCO_3$

Explanation:

Hello,

In this case, since the undergoing chemical reaction is:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The corresponding moles of carbon dioxide occupying 40.0 mL (0.0400 L) are computed by using the ideal gas equation at 273.15 K and 1.00 atm (STP) as follows:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{1.00 atm*0.0400L}{0.082\frac{atm*L}{mol*K}*273.15 K})=1.79x10^{-3} mol CO_2$

Then, since the mole ratio between carbon dioxide and calcium carbonate is 1:1 and the molar mass of the reactant is 100 g/mol, the mass that yields such volume turns out:

$m_{CaCO_3}=1.79x10^{-3}molCO_2*\frac{1molCaCO_3}{1molCO_2} *\frac{100g CaCO_3}{1molCaCO_3}\\ \\m_{CaCO_3}=0.179gCaCO_3$

Regards.

3 0

3 years ago

Which of these statements is true about the temperature of a substance?

Svetradugi [14.3K]

Answer:

(c)

Explanation:

4 0

3 years ago