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2 years ago

What molecule is removed during dehydration synthesis?.

Chemistry

1 answer:

nignag [31]2 years ago

4 0

Answer: <u> two hydrogen atoms and one oxygen</u> atom are removed during dehydration synthesis.

Hope this helps!

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FIRST GETS BRAINLIEST

lisov135 [29]

Answer:

0.07172 L = 7.172 mL.

Explanation:

We can use the general law of ideal gas: <em>PV = nRT. </em>

where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 273 K, Standard T).

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3 years ago

How many electrons are present in 4He2+?

Charra [1.4K]

4 and 2 electrons are present.

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3 years ago

What type of pollution in cities composed of car exhaust and industrial pollutes

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Air pollution is the answer

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2 years ago

Read 2 more answers

Help please! please answer with a good answer

KonstantinChe [14]

Answer:

what the heck what is that?

Explanation:

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2 years ago

Calcule la densidad del hidrógeno H2 en g/L a 327 mm Hg y 48ºc

tankabanditka [31]

<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm , N/m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / MW

m = mass

MW = molecular weight

For density , can be formulated :

$\tt \rho=\dfrac{P\times MW}{R\times T}$

P = 327 mmHg = 0,430263 atm

R = 0.082 L.atm / mol K

T = 48 ºC = 321.15 K

MW of H₂ = 2.015 g/mol

The density :

$\rho=\dfrac{0,430263\times 2.015 }{0.082\times 321.15}\\\\\rho=0.033~g/L$

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2 years ago