What’s my resting energy????

Consider the balanced equation for the following reaction:

Zanzabum

<u>Answer:</u> The amount of carbon dioxide formed in the reaction is 5.663 grams

<u>Explanation:</u>

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 oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol$

For the given chemical equation:

$7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)$

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = $\frac{4}{7}\times 0.25=0.143mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

$0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g$

To calculate the experimental yield of carbon dioxide, we use the equation:

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

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

$90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g$

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

7 0

3 years ago

A gas has a volume of 590 mL at temperature of -55.0 C. What volume will the gas occupy at 30.0 C show your work

DENIUS [597]

Data:
$V_{initial} = 590\:mL$
$T_{initial} = -55.0^0C$
converting to Kelvin
TK = TC + 273
TK = -55.0 + 273 → TK = 218.0 → $T_{initial} = 218.0\:K$
$V_{final} = ? (in\:milliliters)$
$T_{final} = 30.0^0C$
TK = TC + 273
TK = 30.0 + 273 → TK = 303.0 → $T_{final} = 303.0\:K$

By the first Law of Charles and Gay-Lussac, we have:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$

Solving:
$\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }$
$\frac{ 590 }{ 218.0 } = \frac{ V_{f} }{ 303.0 }$
Product of extremes equals product of means:
$218.0* V_{f} = 590*303.0$
$218.0 V_{f} = 178770$
$V_{f} = \frac{178770}{218.0}$
$\boxed{\boxed{V_{f} \approx 820.04\:mL}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago

Igneous rocks would most likely be found near:

geniusboy [140]

Answer:

the deep sea floor. Known as the oceanic crust.

Explanation:

The deep seafloor (the oceanic crust) is made almost entirely of basaltic rocks, with peridotite underneath in the mantle. Basalts are also erupted above the Earth's great subduction zones, either in volcanic island arcs or along the edges of continents.

Hope this helps :)

7 0

3 years ago

_______ is the amount of heat necessary to raise the temperature of 1 gram of a substance by 1 degree Kelvin. A. A calorie B. En

Maksim231197 [3]

<h2>The required option d) "specific heat" is correct.</h2>

Explanation:

To raise the temperature of any substance or material of certain mass to respective temperature it requires some amount of heat.
Specific heat is the amount of heat necessary to raise the temperature of the substance of 1 gram to 1 Kelvin.
It is the amount of heat which is required to raise the temperature per unit mass to per unit temperature.
Thus, the required "option d) specific heat" is correct.

5 0

3 years ago

Calculate the number of sucrose molecules in a 75.0 gram sample.

zzz [600]

Answer:

1.32*10^23 molecules

Explanation:

sucrose formula: C12H22O11

molar mass: 12(12.01)+22(1.01)+11(16.00)=342.34g/mol

75.0 g C12H22O11 * (1 mol C12H22O11)/(342.34g C12H22O11)=0.219 mol C12H22O11

0.219 mol * (6.022*10^23)/mol = 1.32*10^23 molecules (three sig. figures)

5 0

3 years ago