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Sladkaya [172]
4 years ago
13

A chemist prepares a sample of helium gas at a certain pressure, temperature and volume and then removes all but a fourth of the

gas molecules (only a fourth remain). How must the temperature be changed (as a multiple of T1) to keep the pressure and the volume the same?
Chemistry
1 answer:
cupoosta [38]4 years ago
6 0

Answer:

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

Explanation:

Pressure in the container is P and volume is V.

Temperature of the helium gas molecules =T_1

Molecules helium gas = x

Moles of helium has = n_1= \frac{x}{N_A}

PV = nRT (Ideal gas equation)

PV=n_1RT_1...[1]  

After removal of helium gas only a fourth of the gas molecules remains and pressure in the container and volume should remain same.

Molecules of helium left after removal = \frac{x}{4}

Moles of helium has left after removal = n_2= \frac{x}{4\times N_A}

PV=n_2RT_2...[2]

n_1RT_1=n_2RT_2

\frac{x}{N_A}\times T_1=\frac{x}{4\times N_A}\times T_2

T_1=\frac{T_2}{4}

T_2=4T_1

The temperature must be changed to 4 times of the initial temperature so as to keep the pressure and the volume the same.

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How many mole of calcium (Ca) are in 9.00 x 1016 atoms of Ca?
d1i1m1o1n [39]

Answer:

Number mole of calcium (Ca) = 1.4942 x 10⁻⁷ (Approx.)

Explanation:

Given:

Number of atom in (Ca) = 9 x 10¹⁶

Find:

Number mole of calcium (Ca)

Computation:

Number mole of calcium (Ca) = Number of atom in (Ca) / Avogadro's number

Number mole of calcium (Ca) = (9 x 10¹⁶) / (6.023 x 10²³)

Number mole of calcium (Ca) = 1.4942 x 10⁻⁷ (Approx.)

5 0
3 years ago
Calculate the average atomic mass element X
Leto [7]

Answer:

39.02 amu

Explanation:

The average atomic mass of an element can be calculated as follows :

X=\dfrac{9.67\times 38+78.68\times 39+11.34\times 40+0.31\times 41}{100}\\\\X=\dfrac{3902.29}{100}\\\\X=39.02\ amu

So, the atomic mass of the element X is 39.02 amu.

5 0
3 years ago
How much heat is added if 0.814g of water increase in temperature by 0.351 degree C
steposvetlana [31]

Answer:

1·199 J

Explanation:

Given

Mass of water = 0·814 g = 0·814 × 10^{-3} kg

Increase in temperature = 0·351 °C

Let the amount of heat added be Q J

Formula for heat added is

<h3>Q = m × s × ΔT</h3>

where Q is the amount of heat transferred

m is the mass

s is the heat capacity

ΔT is the change in temperature

Heat capacity of water = 4200 J/kg °C

Applying the formula for heat added

Q = 0·814 × 10^{-3} × 4200 × 0·351  = 1·199 J

∴ Amount of heat added = 1·199 J

8 0
4 years ago
Can you guys help me fill out the table?
TEA [102]

Explanation:

Scientist                   Evidence                                 Model

Dalton                        Gases                       indivisible, solid and spheres

J.J Thomson          Deflected beam        Negative charges evenly scattered

                                                                  through positively charged mass of

                                                                  matter.

Rutherford     Deflection of alpha            atomic model

                       particles passing                

                       through the gold foil              

The emboldened words are the answer.

  • John Dalton proposed the first model of the atoms by his works on gases. He postulated the Dalton's law of partial pressure.
  • He suggested that gases are made of tiny particles called atoms.

  • J.J Thomson proposed the plum pudding model of the atom in which the charges are evenly scattered through the positively charged mass of matter.

  • The gold foil experiment by Rutherford brought the atomic model of the atoms in the front-light.
  • The model suggests a small positively charged center which the mass of the atom.
  • The outer space is occupied by the electrons.

Learn more:

Rutherford brainly.com/question/1859083

#learnwithBrainly

6 0
3 years ago
Calculate ΔHo for the following reaction ussing the given bond dissociation energiesCH4(g) + 2O2(g) --&gt; CO2(g) + 2H2O(g)BOND
Mazyrski [523]

Answer:

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

Explanation:

CH_4(g) + 2O_2(g)\rightarrow CO_2(g) + 2H_2O(g) ,ΔH° = ?

We are given with:

\Delta H_{O-O}=142 kJ/mol

\Delta H_{O=O}=498 kJ/mol

\Delta H_{H-O}=459 kJ/mol

\Delta H_{C-H}=411 kJ/mol

\Delta H_{C-O}=358 kJ/mol

\Delta H_{C=O}=799 kJ/mol

ΔH° =  

(Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

\Delta H^o=(1 mol\times 4\times \Delta H_{C-H}+2 mol\times 1\times \Delta H_{O=O})-(1 mol\times 2\times \Delta H_{C=O}+2 mol\times 2\times\Delta H_{H-O})

\Delta H^o=(1 mol\times 4\times 411 kJ/mol+2 mol\times 1\times 498 kJ/mol)-(1 mol\times 2\times 799 kJ/mol+2 mol\times 2\times 459 kJ/mol)

\Delta H^o=-794kJ

\Delta H^o>0 endothermic reaction

\Delta H^o exothermic reaction

The ΔH° for the following reaction is -794 kJ, hence exothermic reaction,

4 0
4 years ago
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