? P+ ? Br→ ? PBT

Under which conditions of temperature and pressure does a real gas behave most like an ideal gas

castortr0y [4]

Answer:

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.

Explanation:

4 0

3 years ago

The molar mass of magnesium (Mg) is 24.30 g/mol. There are 6.02 Times. 1023 atoms in 1 mol.

rjkz [21]

Your answer to this question is 1.20 times 1024 atoms

8 0

2 years ago

When a chemical reaction is run in aqueous solution inside a calorimeter, the temperature change of the water (and Ccal) can be

Yakvenalex [24]

Answer:

The total change in enthalpy for the reaction is - 81533.6 J/mol

Explanation:

Given the data in the question;

Reaction;

HCl + NaOH → NaCl + H₂O

Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J

Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol

so, 0.0500 moles of H₂O produced

Volume of solution = 50.mL + 50.mL = 100.0 mL

Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g

now ,

Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal

we know that; The specific heat of water(H₂O) is 4.18 J/g°C.

so we substitute

q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28

q_soln = 2842.4 + 1234.28

q_soln = 4076.68 J

Enthalpy change for the neutralization is ΔH $_{neutralization}$

ΔH $_{neutralization}$ = -q_soln / mole of water produced

so we substitute

ΔH $_{neutralization}$ = -( 4076.68 J ) / 0.0500 mol

ΔH $_{neutralization}$ = - 81533.6 J/mol

Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol

6 0

2 years ago

Find the pH of the equivalence point and the volume (mL) of 0.0372 M NaOH needed to reach the equivalence point in the titration

elixir [45]

Answer:

8.54

Explanation:

At equivalence point :

42.2 X 0.052 = Vol. NaOH X 0.0372

Vol of NaOH = 2.1944/0.0372 = 58.99 ml

So volume of NaOH recquired to reach equivalence point = 58.99 ml

Number of miliimoles of CH3COOH = molarity X volume in ml = 42.2 X 0.052 = 2.1944 millimoles

Number of millimoles of NaOH = 58.99 X 0.0372 = 2.1944

Now CH₃COOH and NaOH reacts to give CH₃COONa according to the reaction :

CH₃COOH + NaOH ------> CH₃COONa + H₂O

1 mole of CH₃COOH reacts with 1 mole of NaOH to give 1 mole of CH₃COONa

So 2.1944 millimoles of CH₃COOH will react with 2.1944 millimoles of NaOH to give 2.1944 millimoles of CH₃COONa

So all the acid (CH₃COOH) and base (NaOH) has been converted into salt (CH₃COONa) so there is no acid or base left.

Now molarity of CH₃COONa = number of millimoles of CH₃COONa/total volume in ml = 2.1944/(58.99 + 42.2) = 2.1944/101.19 = 0.02169 M

So using the hydrolysis equation :

pH = 1/2 [ pKw + pKa + log c ]

Ka for acetic acid = 1.75 X 10⁻⁵

so pKa = -log (1.75 X 10⁻⁵) = 4.74

Kw = 10⁻¹⁴

so pKw = -log 10⁻¹⁴ = 14

c = 0.02169

so log c = log 0.02169 = -1.66

putting the values....

pH = 1/2 [14 + 4.74 - 1.66 ]

pH = 1/2 [ 17.08] = 8.54

6 0

3 years ago

Read 2 more answers

Find the percent composition of each element in KMnO4

VARVARA [1.3K]

The way you want to find the percent composition would be by breaking down the problem like so:

K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999

Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO

Then you want to take each molar mass and then divide it 110.035 and multiply by 100

Ex. K = 39.098/ 110.035 and the multiply what you get by a 100

You do this for the other elements as well good luck!

6 0

3 years ago