I really need help who ever has the time and skills plzzz help me

A gas at STP occupies 22.4 L if the temperature is changed to 260 K and the pressures changed it to 0.50 ATM what will the new v

asambeis [7]

Answer:

The new volume will be 42, 7 L.

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. The conditions STP are: 1 atm of pressure and 273 K of temperature.

P1xV1/T1 =P2xV2/T2

1 atmx 22,4 L/273K = 0,5atmx V2/260K

V2=((1 atmx 22,4 L/273K )x 260K)/0,5 atm= 42, 67L

3 0

3 years ago

A 100.0 lb skier moves at 40.00miles/hour. Calc her kinetic energy.

Pepsi [2]

answer: its 7290 joules.

explanations: the first procedure is to convert 1 pound to kilogram. 1 kg = 2.205 hence given 100 lb so we cross multiply. 1 kg * 100 = 2.205 * x

hence x= 45 kg. let's convert 1 mile per hour = 0.45 metre per second we cross multiply by 40 mile per hour. x= 40 * 0.45= 18 m/s.

KE= 1/2 * 45 * (18)^2

= 1/2 * 45 * 14580

= 7290joules

5 0

2 years ago

Calculate the density of a substance with a mass of 0.0500g and volume of 6.40mL, round to the correct number of significant num

lozanna [386]

Density=mass/volume
Mass = 0.0500g
Volume = 6.40mL

0.0500g/6.40mL = 0.0078g/mL

5 0

2 years ago

US<br> What is the percent of H in<br> NH3?<br> (H = 1.008 amu, N = 14.01 amu)<br> [?]%

Vlad [161]

Answer:

17.6

Explanation:

Answer is explained above

4 0

2 years ago

Read 2 more answers

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

2 years ago