Ask question

Ask question

All categories

4 years ago

12

which equation is set up correctly to determine the volume of a 1.5 mole sample of oxygen gas at 22 C and 100kPa

1 answer:

Murljashka [212]4 years ago

7 0

Under STP condition, the gas has a rule of 22.4 L per mole. And according to the ideal gas law, V1/T1=V2/T2. Under STP, 1.5 mol gas has volume of 33.6 L. So the volume under 22 C is 33.6*295/273=36.3 L.

You might be interested in

Why is chemistry called the central science

elena55 [62]

Chemistry<span> is often </span>called the central science<span> because of its role in connecting the physical </span>sciences<span>, which include </span>chemistry<span>, with the life </span>sciences<span> and applied </span>sciences<span> such as medicine and engineering.</span>

7 0

3 years ago

Read 2 more answers

The pH of a 0.23M solution of acrylic acid HC3H3CO2 is measured to be 2.44. Calculate the acid dissociation constant Ka of acryl

belka [17]

Answer:

Kₐ = 5.7 x 10⁻⁵

Explanation:

The equilbrium for this acid is

HC₃H₃CO₂ + H2O ⇄ H₃O⁺ + C₃H₃CO₂ ⁻ ,

and the equilibrium constant for acrylic acid is given by the expression:

Kₐ = [ H₃O⁺][ C₃H₃CO₂⁻ ] / [ HC₃H₃CO₂ ]

Since the pH of the 0.23 M solution is known , we can calculate [ H₃O⁺].

The ][ C₃H₃CO₂⁻ ] is equal to [ H₃O⁺] from the above equilibria (1:1)

Finally [ HC₃H₃CO₂ ] is known.

pH = - log [ H₃O⁺]

taking antilog to both sides of the equation

10^-pH = [ H₃O⁺]

Substituting

10^-2.44 = [ H₃O⁺] = 3.6 x 10⁻³

[ C₃H₃CO₂⁻ ] = 3.6 x 10⁻³

Kₐ = ( 3.6 x 10⁻³ ) /0 .23 = 5.7 x 10⁻⁵

7 0

3 years ago

What is the density of a 10 kg mass that occupies 5 liters?<br> ( pls need help)

bulgar [2K]

Answer: d=2000 g/L

Explanation:

Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.

$10kg*\frac{1000g}{1kg} =10000 g$

Now that we have grams, we can divide to get density.

$d=\frac{10000g}{5 L}$

d=2000g/L

7 0

3 years ago

If a laser light is compose of photons each with an energy of 4.58 × 10−19 J, what is the photon energy of this light in units o

kap26 [50]

Answer : The photon energy of this light in units of nanometers, (nm) is, $4.34\times 10^{-2}nm$

Solution :

Formula used :

$E=\frac{h\times c}{\lambda}$

where,

E = energy of photon = $4.58\times 10^{-19}J$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = ?

Now put all the given values in the above formula, we get:

$4.58\times 10^{-19}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}$

$\lambda=4.3017\times 10^{-7}m=434.017\times 10^{-9}m=434.017nm=4.34\times 10^{-2}nm$

conversion used : $(1nm=1\times 10^{-9}m)$

Therefore, the photon energy of this light in units of nanometers, (nm) is, $4.34\times 10^{-2}nm$

5 0

3 years ago

One evening, Alex notes the Moon set in the west at around 8 PM. What would have been the approximate time of moonrise that day?

ra1l [238]

Answer:

C

Explanation:

Early evening about 5 p. m.

6 0

3 years ago

Read 2 more answers