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zepelin [54]
3 years ago
10

Hey sample of helium occupies 235 mL at 788 tour and 25°C. If the sample is condensed into a 0.115 L flask what will be the new

pressure assuming constant temperature?
Chemistry
1 answer:
7nadin3 [17]3 years ago
3 0

the system is in constant temperature

so it follows Boyle's law

i.e.P1V1=P2P2

where P1=initial pressure=788 tour

P2=final pressure= ?

V1=initial volume=235 ml

V2=final volume=0.115 ml

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What is the law of conservation?
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3 years ago
The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J
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Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

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You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

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<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

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<em></em>

I hope it helps!

6 0
2 years ago
La formula quimica de la molecula de agua H20 , si H = 1 gramos y O = 16 gramos . Cual es su composicion porcentual ? 88,88 % de
vesna_86 [32]

Answer:

88,88 % de O y 11,11 % de H

Explanation:

La composición porcentual se define como la masa que hay de cada mol de átomo en 100g. Las moles de agua en 100g son:

<em>Masa molar agua:</em>

2H = 2*1g/mol = 2g/mol

1O = 1*16g/mol = 16g/mol

Masa molar = 2 + 16 = 18g/mol

100g H2O * (1mol / 18g) = 5.556 moles H2O.

Moles de hidrógeno:

5.556 moles H2O * (2mol H / 1mol H2O) = 11.11 moles H

Moles Oxígeno = Moles H2O = 5.556 moles

La masa de hidrógeno es:

11.11mol * (1g/mol) 11.11g H

La masa de oxígeno es:

5.556 mol * (16g / 1mol) = 88.89g O

Así, el porcentaje de O es 88.89% y el de H es 11.11%. La opción correcta es:

<h3>88,88 % de O y 11,11 % de H</h3>
7 0
2 years ago
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