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4 years ago

I only need help in the bottom section? Please someone help me please?

Chemistry

2 answers:

kherson [118]4 years ago

4 0

Energy in the nucleus of an atom of uranium: Nuclear energy
Energy of a moving object: Kinetic Energy
Energy stored in chemical bonds: Potential Energy
Energy emitted from light bulb: Radiant Energy
Energy in a battery: Stored Energy
Energy in our food: Chemical Energy
Energy emitted from a radio: Electromagnetic Energy
Energy affected by mass and speed: Kinetic Energy
Energy affected by position and condition: Potential Energy
Energy from our star that some homes use for electricity: Radiant Energy
I hope this was found helpful!
I was happy to assist you with your Homework :)

jek_recluse [69]4 years ago

3 0

I can't see it can you please send me a picture and I will love to help you.

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3 years ago

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3 years ago

I'LL GIVE U BRAINLIEST

Arturiano [62]

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4 years ago

Read 2 more answers

How many liters of oxygen are required to completely react with 2.0 liters of CH4 at30 °C and 3.0 atm?CH4(g) + 2O2(g) → CO2(g) +

Dominik [7]

1) Write the chemical equation.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

2) List the known and unknown quantities.

Sample: CH4.

Volume: 2.0 L.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Moles: unknown.

3) Moles of CH4.

3.1- Set the equation.

$PV=nRT$

3.2- Plug in the known values and solve for n (moles).

$(3.0\text{ }atm)(2.0\text{ }L)=n*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $n=\frac{(3.0\text{ }atm)(2.0\text{ }L)}{(0.082057\text{ }L*atm*K^{-1}*mol^{-1})}=$ $n=0.24\text{ }mol\text{ }CH_4$

4) Moles of oxygen that reacted.

The molar ratio between CH4 and O2 is 1 mol CH4: 2 mol O2.

$mol\text{ }O_2=0.24\text{ }CH_4*\frac{2\text{ }mol\text{ }O_2}{1\text{ }mol\text{ }CH_4}=0.48\text{ }mol\text{ }O_2$

5) Volume of oxygen required.

Sample: O2.

Moles: 0.48 mol.

Temperature: 30 ºC = 303.15 K.

Pressure: 3.0 atm.

Ideal gas constant: 0.082057 L * atm * K^(-1) * mol^(-1).

Volume: unknown.

5.1- Set the equation.

$PV=nRT$

5.2- Plug in the known values and solve for V (liters).

$(3.0\text{ }atm)(V)=0.48\text{ }O_2*(0.082057\text{ }L*atm*K^{-1}mol^{-1})(303.15\text{ }K)$ $V=\frac{(0.48\text{ }mol\text{ }O_2)(0.082057\text{ }L*atm*K^{-1}*mol^{-1})(303.15\text{ }K)}{3.0\text{ }atm}$ $V=3.98\text{ }L$

3.98 L of O2 is required to react with 2.0 L CH4.

6 0

2 years ago

a certain anesthetic contains 64.9% C, 13.5% H, and 21.6% O by mass. at 120 deg Celsius & 750 mmHg, 1.00 L of the gaseous co

slamgirl [31]

You need to use the % information to determine the empirical formula of the compound first.

The empirical formula is the simplest ratio of atoms in the molecule.

Then use the rest of the data to determine moles of gas, and use this to determine molar mass of gas...

Empirical formula calculations:

Assume you have 100 g, calculate the moles of each atom in the 100 g

moles = mass / molar mass
molar mass C = 12.01 g/mol
molar mass H = 1.008 g/mol
molar mass O = 16.00 g/mol

C = 64.9 % = 64.6 g
H = 13.5 % = 13.5 g
O = 21.6 % = 21.6 g

moles C = 64.6 g / 12.01 g/mol = 5.38 mol
moles H = 13.5 g / 1.008 g/mol = 13.39 mol
moles O = 21.6 g / 16.00 g/mol = 1.35 mol

So ratio of C : H : O
is 5.38 mol : 13.39 mol : 1.35 mol

Divide each number in the ratio by the lowest number to get the simplest whole number ratio

(5.38 / 1.35) : (13.39 / 1.35) : (1.35 / 1.35)

4 : 10 : 1

empirical formula is
C4H10O

Finding moles and molar mass calcs

Now, you know that at 120 deg C and 750 mmHg that 1.00L compound weighs 2.30 g.

We can use this information to determine the molar mass of the gas after first working out how many moles the are in the 1.00 L

PV = nRT
P = pressure = 750 mmHg
V = volume = 1.00 L
n = moles (unknown)
T = temp in Kelvin (120 deg C = (273.15 + 120) Kelvin)
- T = 393.15 Kelvin
R = gas constant, which is 62.363 mmHg L K^-1 mol^-1 (when your P is in mmHg and volume is in L)

n = PV / RT
n = (750 mmHg x 1.00 L) / (62.363mmHg L K^-1 mol^-1 x 393.15 K)
n = 0.03059 moles of gas

We know moles = 0.03509 and mass = 2.30 g
So we can work out molar mass of the gas

moles = mass / molar mass
Therefore molar mass = mass / moles
molar mass = 2.30 g / 0.03059 mol
= 75.19 g/mol

Determine molecular formula

So empirical formula is C4H10O
molar mass = 75.19 g/mol

To find the molecular formula you divide the molar mass by the formula weight of the empirical formula...
This tells you how many times the empirical formula fits into the molecular formula. Tou then multiply every atom in the empirical formula by this number

formula weight C4H10O = 74.12 g/mol

Divide molar mass by formula weight empirical
75.15 g/mol / 74.12 g/mol
= 1
(It doesn't matter that the number don't quite match, they rarely do in this type of calc (although I could have made a slight error somewhere) but the numbers are very close, so we can say 1.)

The empirical formula only fits into the molar mass once,

molecular formula thus = empirical formula

C4H10O

Therefore, the molecular formula of the compound is C4H10O.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

5 0

3 years ago