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3 years ago

The conversion of mass into huge amounts of energy is the basis of __________.

Chemistry

2 answers:

Illusion [34]3 years ago

7 0

The Answer is D, nuclear energy

pshichka [43]3 years ago

4 0

Answer:

Explanation:

Nuclear binding energy curve. During the nuclear splitting or nuclear fusion, some of the mass of the nucleus gets converted into huge amounts of energy and thus this massis removed from the total mass of the original particles, and the mass is missing in the resulting nucleus.

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A runner competed in a 5-mile run. How many yards did she run?

vfiekz [6]

5 miles are equivalent to 8,800 yards

so b. is the correct answer. plz like and hope it helped

3 0

3 years ago

Take car traveling with constant speed travels at 150 km in 7200 a. what is the speed of the car?

Andrej [43]

<span>Find the speed of the car in which it travels at 150 km in 7200 seconds.
This is an easy question, you just need to follow the given formula as always.
Since the time and the distance is already given, we are now looking for the speed
=> time = 7200 seconds
=> distance = 150km</span><span>
d = speed x time
S = distance / time
s = 150 km / 7200 seconds
s = 0.021 km / seconds.
Thus, the car travels for about 0.021 km per seconds.

</span>

5 0

3 years ago

Suppose 6.63g of zinc bromide is dissolved in 100.mL of a 0.60 M aqueous solution of potassium carbonate. Calculate the final mo

Alenkasestr [34]

Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

1.46x10⁻¹⁰ = [Zn²⁺] [CO₃²⁻]

We can find the moles of ZnBr₂ added = Moles of Zn²⁺ and moles of K₂CO₃ = Moles of CO₃²⁻ to find the moles of CO₃²⁻ that remains in solution, thus:

<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

4 0

3 years ago

Name three sources of atmospheric carbon dioxide.

timurjin [86]

Answer:

Natural sources include decomposition, ocean release and respiration. Human sources come from activities like cement production, deforestation as well as the burning of fossil fuels like coal, oil and natural gas.

4 0

3 years ago

Methanol has a normal boiling point of 64.6C and a heat of vaporization of 35.2 kJ/mol. What is the vapor pressure (in Torr) of

DENIUS [597]

Answer:

vapor pressure of methanol at 12.0C = 75.09 torr

Explanation:

Using Clausius Clapeyron equation

, we have that

ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)

Given

At Normal boiling point,

Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm

Heat of vaporization = 35.2 kJ/mol

Changing to J/mol

=35.2 x 1000= 35200 J/mol

Temperature , T2 = 12.0oC = 12 + 273 = 285 K

Using gas constant, R = 8.314 J/mol.K

ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)

ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)

ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)

ln (P2/ 1 atm) = 4,233.822468 x-0.0005466866

ln (P2/ 1 atm)= -2.31457

P2 = e^⁻2.31457 x 1 atm

P2=0.098808atm

= 0.098808atm x760 = 75.09 torr

7 0

3 years ago