Ask question

Ask question

All categories

2 years ago

5

The vapor pressure of ethanol at 25 Celsius is 0.0773 atm. Calculate the vapor pressure in mmHg and torr. Round each of your ans

wers to 3 significant digits.

1 answer:

Elden [556K]2 years ago

8 0

Answer:

hope it helps!Brainliest pls

Explanation:

You might be interested in

8. What is probability?

nata0808 [166]

Answer:

<em>In equipment reliability, the likelihood that a given event will occur</em>

Explanation:

6 0

3 years ago

A sample of oxygen gas has a volume of 150.0 ml when its pressure is .947 atm. what will the volume of the gas be if the tempera

joja [24]

D
Hope it helped you get an A, Kayla.

7 0

2 years ago

If a 125 gms of radioactive element has a half life of 60 min how many half lives will it go through to become 3.90625 gms

Sonja [21]

Answer:

5

Explanation:

Let the number of half lives be x

<u>Solve this equation to find the value of x:</u>

125*(1/2)ˣ = 3.90625
(0.5)ˣ = 3.90625 / 125
(0.5)ˣ = 0.03125
log (0.5)ˣ = log 0.03125
x = log 0.03125 / log 0.5
x = 5

4 0

2 years ago

antimony has two naturally occuring isotopes, sb121sb121 and sb123sb123 . sb121sb121 has an atomic mass of 120.9038 u120.9038 u

Luda [366]

Considering the definition of atomic mass, isotopes and atomic mass of an element, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

<h3>Definition of atomic mass</h3>

The atomic mass is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

<h3>Definition of isotope</h3>

Isotopes are the chemical elements in which atomic numbers are the same, but the number of neutrons is different.

<h3>Definition of atomic mass</h3>

The atomic mass of an element is the weighted average mass of its natural isotopes.

This is, the atomic masses of elements are usually calculated as the weighted average of the masses of the different isotopes of each element, considering the relative abundance of each of them.

<h3>Percent natural abundance of each isotope</h3>

In this case, antimony has two naturally occuring isotopes, sb121 and sb123. You know:

sb121 has an atomic mass of 120.9038 u.
sb121 has a percent natural abundance of x.
sb123 has an atomic mass of 122.9042 u.
sb123 has a percent natural abundance of 1 -x (or, what is the same, the abundance is 100% - x%, since both isotopes form 100% of the element.)
Antimony has an average atomic mass of 121.7601 u

The average mass of antimony is expressed as:

121.7601 u= 120.9038 u x + 122.9042 u× (1 -x)

Solving:

121.7601 u= 120.9038 u x + 122.9042 u - 122.9042 u x

121.7601 u - 122.9042 u= 120.9038 u x - 122.9042 u x

(-1.1441 u)= (-2.0014) x

(-1.1441 u)÷ (-2.0014)= x

<u><em>0.5726= x or 57.26%</em></u>

So, 1 -x= 1- 0.5716 → <u><em>1-x= 0.4284 or 42.84%</em></u>

<u><em /></u>

Finally, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

Learn more about average atomic mass:

brainly.com/question/4923781

brainly.com/question/1826476

brainly.com/question/15230683

brainly.com/question/7955048

#SPJ1

5 0

1 year ago

1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

<em>Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:</em>

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

<em>Q2: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation </em>

<em>qwater = m × c × ΔT. </em>

<em>We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.</em>

<em></em>

a) First part: the energy change (q) of the surroundings (water):

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

<em>∴ The amount of heat absorbed by water = Q = m.c.ΔT</em> = (25 g)(4.18 J/g°C)(6.4°C) = <em>668.8 J.</em>

<em>b) second part:</em>

<em>Q water = Q unknown metal. </em>

<em>Q unknown metal = - </em>668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

<em>Q unknown metal = - </em>668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

<em>- </em>668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (<em>- </em>668.8)/(- 1914) = 0.3495 J/g°C.

<em></em>

3 0

3 years ago