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3 years ago

14

If 2.40 moles of gas are held at a temperature of 97.0 °C in a container with a volume of 45.0 L, what is the pressure of the ga

s?

1 answer:

Rainbow [258]3 years ago

7 0

PV=nRT<=> P=nRT/V=2,40*R*(273+97)/45 atm.
Calculate it. R is a number that is given, find it and use your math to solve.

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A sample of gas has an initial volume of 3.62 L at a pressure of 0.987 atm. If the volume of the gas decreases to 2.50 L, what w

statuscvo [17]

Answer:

0.033 atm

Explanation:

PV=nRT

rule of 1 for T and n

P= nRT/V

P= 1(.08206)1)/ 2.50

0.033 atm = pressure

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3 years ago

Which statement describes J.J. Thomson’s plum-pudding model of the atom?

ra1l [238]

Option C: the electrons are embedded in a positive sphere.

As we know that was a wrong model and nowadays the electrons are considered to be around the nucleus, in regions called orbitals.

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4 years ago

At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival

evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0

3 years ago

Read 2 more answers

Which of the following is NOT an advantage of the metric system?

notsponge [240]

Answer: The answer is A

Explanation:

7 0

3 years ago

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A solution contains one or more of the following ions: Ag , Ca2 , and Co2 . Lithium bromide is added to the solution and no prec

koban [17]

Answer:

$Ca^{2+} and Co^{2+}$

Explanation:

Given:

A solution contains one or more of the following ions such as Ag, $Ca_2$ and $Co_2$

Here the Lithium bromide is added to the solution and no precipitate forms

Solution:

Since with LiBr no precipitation takes place therefore Ag+ is absent

Here on adding $Li_2SO_4$ to it precipitation takes place.

Precipitate is as follows,

$Ca^{2+}(aq)+SO_4^2^-(aq)----->CASO_4(s)$

Thus,

$Ca^2^+$ is present

When $Li_3PO_4$ is added again precipitation takes place.

Therefore the reaction is as follows,

$Co^2^+(aq)+PO_4^3^-(aq)------>Co_3(PO_4)_2(s)$

Therefore,

$Ca^{2+} and Co^{2+}$ are present in the solution

3 0

3 years ago