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2 years ago

Write the chemical equation for cellular respiration

Chemistry

1 answer:

BaLLatris [955]2 years ago

7 0

Answer:

C 6 H 12 O 6 + 6O 2 → 6CO 2 + 6H 2 O + 38 ATP

Explanation:

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¿Cuál es la diferencia entre los compuestos alifáticos y ciclicos?

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2 years ago

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Which of these is a characteristic of pseudoscience but not a characteristic of science? (4 points)

Galina-37 [17]

Answer:

The Answer is B

Explanation:

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2 years ago

The length of a sheet of U.S standard (letter size) paper is closest to: 8 centimeters 11 centimeters 29 centimeter 300 centimet

s2008m [1.1K]

US Letter is a paper size, which measures:
8.5 inches ( width ) by 11.0 inches ( length ).
8.5 inches = 21.59 cm
11.0 inches = 27.94 cm
The length ( 27.94 cm ) is closest to 29 cm.
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2 years ago

If you are given an ideal gas with pressure (p)259,392.00 pa and temperature (T)=200°c of 1 mole Argon gas in a volume 8.8dm3,ca

GuDViN [60]

Answer: $R=4.82436 \frac{Pa. m^{3}}{mol. K}$

Explanation:

The Ideal Gas equation is:

$P.V=n.R.T$ (1)

Where:

$P$ is the pressure of the gas

$n$ the number of moles of gas

$R=8.3144598 \frac{Pa. m^{3}}{mol. K}$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin.

$V$ is the volume

It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that <u>an ideal gas is a single hypothetical gas</u>. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be $8.3144598 \frac{Pa. m^{3}}{mol. K}$ .

However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.

Having this clarified, let's isolate $R$ from (1):

$R=\frac{PV}{nT}$ (2)

Where:

$P=259392 Pa$

$n=1 mole$

$T=200\°C=473.15 K$ is the absolute temperature of the gas in Kelvin.

$V=8.8 dm^{3}=0.0088 m^{3}$

$R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}$ (3)

Finally:

$R=4.82436 \frac{Pa. m^{3}}{mol. K}$

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2 years ago

Using the periodic table, how would you find elements with chemical properties similar to helium?

rosijanka [135]

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3 years ago