All categories

2 years ago

8. A party-supply store sells a helium-gas

Chemistry

1 answer:

MatroZZZ [7]2 years ago

7 0

Answer:

P₂= 116.7 atm

Explanation:

Here apply the Boyle's law equations that states :at constant temperature the volume of a dry mass of a gas is inversely proportional to its pressure.

This is simplified as;

P₁V₁=P₂V₂ where P is pressure and V is volume

Given that;

P₁=1

V₁=1.81 m³

P₂=?

V₂=1.55*10^-2 m³

Apply the formula

1*1.81 =P₂*1.55*10^-2 m³

1.81/1.55*10^-2 =P₂

P₂= 116.7 atm

You might be interested in

How does a barometer predict the weather?

Lunna [17]

Answer:

Barometers measure this pressure. ... Changes in the atmosphere, including changes in air pressure, affect the weather. Meteorologists use barometers to predict short-term changes in the weather. A rapid drop in atmospheric pressure means that a low-pressure system is arriving.

Explanation:

<h3>I hope this helps!</h3>

6 0

2 years ago

Which models of the atom in task 1 are not supported by the results of the hydrogen gas experiment? For each of these models, ex

kramer

Answer:

Thomson placed two magnets on either side of the tube, and observed that this magnetic field also deflected the cathode ray. The results of these experiments helped Thomson determine the mass-to-charge ratio of the cathode ray particles, which led to a fascinating discovery, minus the mass of each particle was much, much smaller than that of any known atom. Thomson repeated his experiments using different metals as electrode materials, and found that the properties of the cathode ray remained constant no matter what cathode material they originated from. From this evidence, Thomson made the following conclusions:

The cathode ray is composed of negatively-charged particles.

The particles must exist as part of the atom, since the mass of each particle is only ~1/2000 the mass of a hydrogen atom.

These subatomic particles can be found within atoms of all elements.

While controversial at first, Thomson's discoveries were gradually accepted by scientists. Eventually, his cathode ray particles were given a more familiar name: electrons. The discovery of the electron disproved the part of Dalton's atomic theory that assumed atoms were indivisible. In order to account for the existence of the electrons, an entirely new atomic model was needed.

Explanation:

8 0

1 year ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$