A sample of gas in a balloon has an initial temperature of 2.0 ∘C and a volume of 1.53×103 L . If the temperature changes to 99 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

Answers

Answer 1
Answer: (V1/T1)=(V2/T2) this is Charles' law where T is in Kelvin
So: (1.53x10^3)/(275K) = V2/(372K)
V2= 2.07x10^3L

Related Questions

Which two properties are used to define matter? A.has mass and conducts heat B.has buoyancy and volume C.has mass and takes up space D.conducts energy and has characteristic properties
the balanced chemical equation for one neutralization reaction. In your equation, which reactant contributed the salt's positive ion? Which one contributes the salt's negative ion?
Which of these statements about enzymes is NOT true? - If enough substrate is added, the normal Vmax of a reaction can be attained even in the presence of a competitive inhibitor. - When [S] << Km, the reaction is second order and V0 depends on [S] and [Et]. - Their kcat is a second order rate constant. - The lower their Km, the better they recognize their substrate, but the lower their reaction rate. - When [S] << Km, V0 depends on [S] and [Et].
How do you think you can use the calorimeter to compare the specific heat capacities of a substance?
In Experiment 1, the density of pure ethanol was found to be:Question 1 options:less than 0.793 g/mL.0.793 g/mL.0.999 g/mL.greater than 0.999 g/mL.

Every atom of the _______ carbon has 6 protons.

Answers

The answer is element. Elements consist of identical atoms, which differ between elements. Atoms are the basic units of matter. They are made up of protons, neutrons, and electrons. Carbon is an element. It contains only of carbon atoms. Every atom of element carbon consists of 6 protons.Hope this helps. Let me know if you need additional help!

Answer: Element

Explanation: hope this helps :3

Which element in the alkali group will have the lowest first ionization energy?

Answers

Fluorine

Explanation:

F- has the lowest ionisation energy

When star A runs out of fuel, it explodes in a supernova and becomes a neutron star. When star B runs out of fuel, it contracts to become a white dwarf. Based on this information which of the following conclusions is most likely true:A. Star A had a larger mass than star B
B. For most of its life cycle, star A appeared substantially brighter than star B from earth
C. For most of its life cycle, star B appeared substantially brighter than star A from earth

Answers

Only five supernovae have ever been documented as being visible to the unaided eye, with two of those believed to be core-collapse supernovae. Thus, option B is correct.

What is are the process of supernova formation?

Iron atoms are compressed so closely together that the nuclei's repelling forces cause the squeezed core to recoil, which causes the star to go supernova and generate a massive, superheated shock wave.

A supernova explosion takes place in a star over the course of just 100 seconds. A star that experiences a supernova explosion dies and leaves behind a neutron star or a black hole as a remnant.

The largest explosion that mankind have yet witnessed is a supernova. Each explosion is the incredibly bright, incredibly potent explosion of a star. The largest explosion that mankind have yet witnessed is a supernova.

Therefore, For most of its life cycle, star A appeared substantially brighter than star B from earth

Learn more about supernova here:

brainly.com/question/28348110

#SPJ2

B. for most of its life cycle, star A appeared substantially brighter than star B from earth

Which best describes the purpose of a controlled experiment?

Answers

A controlled experiment is an experimental setup designed to test the hypotheses. It has one or more conditions (independent variables) and measures (dependent variables). It is necessary to change more than one variable, but all os the experimental conditions will be controlled so that only the variables being examined change and the amount or way of change is measured.

As a radioactive element emits gamma radiation only, the atomic number of the element increases or decreases

Answers

Radioactive elements are the unstable nuclei of the atoms that release energy and radiation. The atomic number remains the same in case of the gamma radiation.

What is gamma decay?

Gamma radiations are the particles that are released when the unstable atom undergoes decay and emits charged particles. These rays have no effect on the atomic mass and number of the atom.

These rays only emitenergy whereas alpha and beta decay produces energy and newparticle. An excited nucleus emits energy in the case of gamma decay that is unlike the other two decays.

Therefore, there is no change in the atomicnumber in gamma decay.

Learn more about gamma decay here:

brainly.com/question/14087656

#SPJ2

Answer: It actually remains the same.

Explanation: Gamma radiation does not affect the atomic number or the atomic mass.

A closed system initially containing 1×10^-3M H2 and 2×10^-3 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture show s that the concentration of HI is 1.87×10^ -3. Calculate Kc at 448°C for the reaction.

Answers

Answer:

1.74845

Explanation:

We have the following reaction:

I2 + H2 => 2 HI

Now, the constant Kc, has the following formula:

Kc = [C] ^ c * [D] ^ d / [A] ^ a * [B] ^ b

In this case I2 is A, H2 is B and C is HI

We know that the values are:

 H2 = 1 × 10 ^ -3 at 448 ° C

I2 = 2 × 10 ^ -3 at 448 ° C

HI = 1.87 × 10 ^ -3 at 448 ° C

Replacing:

Kc = [1.87 × 10 ^ -3] ^ 2 / {[2 × 10 ^ -3] ^ 1 * [1 × 10 ^ -3] ^ 1}

Kc = 1.87 ^ 2/2 * 1

Kc = 1.74845

Which means that at 448 ° C, Kc is equal to 1.74845

Answer:

K_c = 51

Explanation:

[H2] = 10^-3

[I2] = 2*10^-3

[HI] = 0

in equilbiirum

[H2] = 10^-3 - x

[I2] = 2*10^-3 -x

[HI] = 0 + 2x

and we know

[HI] = 0 + 2x = 1.87*10^-3

x = ( 1.87*10^-3)/2 =  0.000935

then

[H2] = 10^-3 - 0.000935 = 0.000065

[I2] = 2*10^-3 -0.000935 = 0.001065

                        H₂                +          I              ⇄              2 HI

Initially     1 × 10⁻³                       2 × 10⁻³

Change  -9.35 × 10⁻⁴               -9.35 × 10⁻⁴                +1.87 × 10⁻³

At equil   6.5 × 10⁻⁵                  1.06 5 × 10⁻³               1.87 × 10⁻³

HI increase by 1.87 × 10⁻³M

K_c = ([HI]^2)/([H_2][I_2]) \n\n= ((1.87*10^-^3)^2)/((6.5*10^-^5)(1.065*10^-3)) \n\nK_c = 51