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Mice21 [21]
3 years ago
11

Help Please and answer correctly otherwise, I can report the answer and your account so make sure the answer is correct BUT PLEA

SE HELP ME I AM BEGGING. BRAINLY PLEASE PUT ME AT THE TOP OF THE PAGE PLEASE.

Chemistry
1 answer:
Alina [70]3 years ago
8 0

Answer:

G

Explanation:

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barxatty [35]
The frequency stays the same it just gets louder
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3 years ago
Which postmortem parameter reaches a maximum at 12 hours and is gone within 36 to 48 hours?1.rigor mortis2.cloudiness of the eye
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3 0
3 years ago
Need help !!!!! ASAP
Ksivusya [100]
<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of CO_{2}

n=0.120mol

<h2>Why?</h2>

We are asked to calculate the number of moles of the given gas, also, we  are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

PV=nRT

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

R=62.363\frac{mmHg.L}{mol.K}

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

Temperature(K)=Temperature(C\°) + 273K

So, we are given the following information:

Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K

Now, isolating the number of moles, and substituting the given information, we have:

PV=nRT

n=\frac{PV}{RT}

n=\frac{PV}{RT}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole

Hence, we have that there were produced 0.120 moles of CO_{2}

n=0.120mol

Have a nice day!

7 0
3 years ago
How many L of a 7.5 H2SO4 stock solution would you need to prepare a dilute solution 100 L of 0.25M H2SO4
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Answer:

3.33 L

Explanation:

We can solve this problem by using the equation:

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Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:

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Where does the rest of the energy “lost” go? <br> (In food webs)
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3 0
3 years ago
Read 2 more answers
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