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exis [7]
2 years ago
15

Name the laws for 1,2, and 3

Chemistry
1 answer:
BaLLatris [955]2 years ago
7 0

Answer:

1. Boyle's law

2. Charle's law

3. Ideal Gas law

Explanation:

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Check all statements below that are true.
docker41 [41]

Answer:

The answers are A,B,C.

Explanation: Just got it right on Edge 2020

3 0
3 years ago
A 5 mL sample of an unknown liquid is placed into a 250 mL flask that has had all of the air removed. The pressure measured in t
Brrunno [24]

Answer:

See explanation below

Explanation:

In this case, we can use the Boyle's law. Assuming that the temperature of both trial remains constant, then:

P₁V₁ = P₂V₂    (1)

You should note that this expression is usable when you are dealing with gases. However, we can treat this unknown liquid as a gas, because all the air on the flask is removed, and we can assume that the liquid can behave like an ideal gas.

So using the above expression, we can solve for P₂:

P₂ = P₁V₁ / V₂    (2)

In this case, we already have the values of presures and volume, so replacing in this expression:

P₂ = 34.5 * 5 / 214

<h2>P₂ = 0.806 kPa</h2>

This should be the pressure of the liquid.

Hope this helps

8 0
2 years ago
Can i get this answer pls
RSB [31]

Answer:

A

Explanation:

In genetic terms, a dominant trait is one that is phenotypically expressed in heterozygotes

3 0
2 years ago
Read 2 more answers
Given R
icang [17]

Answer:

Explanation:

there are  a certain amount of atoms

6 0
2 years ago
El pH de una solución acuosa 10^-14 M de ácido acético, a 25°C, es igual a, teniendo en cuenta que Ke = 1.76x10^-5
frez [133]

Answer:

pH=14

Explanation:

Hola!

En este caso, consideramos que la disociación de ácido acético ocurre:

CH_3COOH\rightarrow CH_3COO^-+H^+

Así, mediante la solución del equilibrio ácido, podemos calcular la concentración de iones hidronio que posteriormente sirven para calcular el pH de la solución, por tal razón, debemos calcular el equilibrio dada la constante de equilibrio y por medio de la ley de acción de masas en términos del cambio x como cualquier problema de equilibrio:

Ke=\frac{CH_3COO^-][H^+]}{[CH_3COOH]}\\\\1.76x10^{-5}=\frac{x*x}{1x10^{-14}M-x}

Resolviendo para x, tenemos x=0.999x10^{-14}

Así, la concentración de hidrógeno es igual a x, por lo que el pH:

pH=-log([H^+])=-log(0.999x10^{-14})\\\\pH=14

Dicho valor tiene sentido desde que la concentración de hidrógeno es casi despreciable, por lo que se puede asumir que tiende a ser básica.

Saludos!

5 0
3 years ago
Read 2 more answers
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