if a liter of water is heated from 20ºc to 50ºc, what happens to its volume?

give the temperature and pressure at stp.

 The stp is the standard temperature and pressure, the standard temperature is o celcius , 273 kelvin and the standard pressure is 1 atmosphere

a gas occupies 4.23 l at 2.25 atm. what is the volume at 3.46 atm?

 According to boyle’s law pressure is inversely proportional to volume , when volume increases, the pressure reduces

 As the pressure increases from 2.25 atm to 3.46 atm , the volume decreases from 4.23 l to 2.75l,

 Boyles law states that the volume of a fixed mass of a gas is inversely proportional to its absolute pressure provide the temperature remain constant.

if the temperature of a piece of steel decreases what happens to its density

if a temperature of a piece of still decreases, the volume also decreases, we know that according to the Charles law that temperature is directly proportional to the volume provided that the pressure remains constant

if the temperature of the steal increases the volume also increases and if the temperature decreases the volume also decreases.

 Therefore if the temperature of an object increases it volume also increases .

 But density is  given as

 Density is given as mass per unit volume,

 Density is inversely proportional to volume  and likewise the temperature, therefore when the temperature increases, the volume expand and this reduces the mass to volume ratio, which is temperature.

 We can say this in other words, when the temperature of a body increases, it atoms and molecules change more kinetic energy, the  move and expand faster ( depending on the state of the matter). The increases in volume decreases the mass per unit volume ratio which is density

as the density of a substance increases, the volume of a given mass of that substance  also increases.

if a liter of water is heated from 20ºc to 50ºc, what happens to its volume?

 when a liter of water is heater from 20 Celsius to 50 celsius,  the temperature increases and according to  the Charles law,

 Charles the temperature of a fixed mass of a gas is directly proportional to its volume provided pressure remain constant.

 If the temperature of a gas increases the volume will also increases.

the density of ethanol is 0.789 g/cm3 at 20°c. what volume of ethanol has a mass of 15.5 grams?

Density is given as

VOLYME= 0.789* 15.5

= 12.2295 CM CUBE

when oil spills in the ocean why does the oil float on the water

 when oil spill in the ocean , the oil will float on top of the water because the oil less dense than the water.

 The water with density of 1g per cm cube is denser than oil. Therefore the heavy mass will sink will the less dense will float.

how to clean oil spills in water

 we can clean oil spill on water by using what will break the oil molecules away.

This include using detergent, chemicals, booms and

most substances become less dense as the temperature increases which substance is an exception

 water is the exception, water at 4 degree centigrade tend to behave abnormally when heated, the volume decreases instead of expanding, tis is called anomalous expansion of water.

 When water is at temperature below zero degree centigrade, the volume increases as the temperature , reaches  zero degree it tends to decreases n volume as the temperature continue to increase until it reaches 4 degree celsius

what is the density of an object having a mass of 8.0 g and a volume of 25 cm3?

when oil spills in the ocean, why does the oil float on the water?

 The oil float because the oil is less dense than water, when two liquid with different densities are mixed the denser liquid will sink while the lighter liquid will float.

the work done in pushing a tv set a distance of 2 m with an average force of 20 n is

workdone= force * distance,

 the force is given as  20 newton and the distance moves is 2 meter

 therefore the workdone = 20*2=40joules.

the power required to exert 4-n force over 3 meters in 2 seconds is

 power is rate of energy consumption

 or power is  workdone divided by time

workdone = 4* 3= 12 Newton

t =2seconds

 power = 12/2 = 6j/s

when a car is braked to a stop, its kinetic energy is transformed to

 when  car is break to stop it kinetic energy is converted to the frictional force needed to top the, the car, also part of it is converted to heat energy.

 We cannot say that it is converted to potential energy but a car moving has a kinetic energy and by the time it stop it loses it mechanical energy.

the mass of a ball moving at 3 m/s with a momentum of 48 kg m/s is

momentum = mass * velocity

 =mv

48= m*3

M=48/3

M=16kg

a rifle recoils while firing a bullet. the speed of the rifle's recoil is small because the

the speed of the rifle recoil is small compared to that of the bullet because , the bullet has a very small mass compared to the rifle.

 And the newton’s law of motion requires that  the action and reaction must  be equal, thereforefore the recoil  velocity mut be small since the mass is large and on the bullet side the recoil velocity must be very high since the bullet mass is very small.

Mv =mu

“ to every action there is an equal and opposite reaction”

a ball rolling down an incline has its maximum kinetic energy at

 a ball  rolling down an incline plane has it maximum kinetic energy  when the velocity is maximum, as a ball start to roll down an inclined plane it acceleration  changes with time, meaning it velocity increases and this velocity is maximum just before it hit the ground.

When a ball start to roll , it potential energy is maximum while the kinetic energy is minimum, when it is midway the potential and kinetic energy are equal, when it is just about to hit the ground the potential energy is minimum while the kinetic energy is maximum

a feather and a coin dropped in a vacuum fall with equal

 a feather and coil falling through a vacuum will have same acceleration due to gravity  due to the absent of air resistance.

 In the absent of air resistance all object will fall with same acceleration due to gravity.

the unit kilowatt-hour is a unit of

is the unit for measuring electric power consumption.

an object has gravitational potential energy due to its

mass

binding energy per nucleon of helium-4

to find the binding energy per  nucleon for helium nucleus, we first have to find the binding energy for the helium atom and then divide it by the mass number of elium atom

 the binding energy is given by

B.E= mc²

Where mass is the mass defects, and c is the speed of light in 3*10⁸m/s

The mass defects is the difference between the calculated mass and the measured mass of an atom.

Helium has a mass of 4.002602 u

To find the calculated mass

 We start by

Helium-4 as an  mass number of 4 and atomic number of 2

 Therefore N= A-Z

N=4-2 =2

 Therefore   N(1.008664 u) +Z(1.00727647)

 2.017328 + 2.014552= 4.03188

Mass defects = calculated mass –measured  mass

= 0.029278

 This the mass defect,

B.E=  0.029278 *  3*10⁸m/s

=8783400 * 1.67*10^-27kg

= 0.000000000000000000014668278joules

[h1]calculate the binding energy eb of the nitrogen nucleus 14 7n.

[/h1]

 to calculate the binding energy of nitrogen, let us find the

 mass defect of nitrogen-14

 nitrogen-14 has an atomic number of  7 and mass number of 14

 the mass defect is the difference between the calculated mass and the measured mass,

 the calculated mass can be obtained as follows

Therefore   N(1.008664 u) +Z(1.00727647)

7(1.008664 u) +7(1.00727647)

7.060648 + 7.05093529

=14.11158329AMU

  The mass defects is = calculated mass-  measured  mass

Measured mass = 14.003241

The mass defects is = calculated mass-  measured  mass

14.11158329- 14.003241= 0.10834229amu

Mass defect is 0.10834229

B.E = mc²

But mass is measured in kg, therefore we have to convert our mass defects to  kg

0.10834229 * mass of electron

0.10834229* 1.67* 10²⁷kg

1.809316243e*10^-28 kg

B.E =MC²

= 1.809316243e*10^-28 kg *(3*10⁸)²

1.809316243e*10^-28 kg *(9*10¹⁶)

=0.000000000016283846187 JOULES

Solve 3215p→ 0−1e+?: physics tutorial

Solve 3215p→ 0−1e+?

 To  solve the above equation, let me rewrite the equation

³²₁₅P→ ⁰₋₁e+?

we can balance , the above equation, by using the law of conservation of masses and charges, the law requires all mass number on both side must be equal and all atomic number for both side must be equal

let me rewrite the above equation

³²₁₅P→ ⁰₋₁e+^a_zX

Where a I the mass number, z is the atomic number and X is the unknow nuclide

For the mass number

We can have

32= 0+a

32-0=a

32=a

a =32

and for the atomic number.

15=(-1)+z

15+1 =z

z= 16

 therefore the nuclide formed has an atomic number of 16 and a mass number of 32.

 The atom with the above property is sulphur-32

 The above equation, will then be written as

³²₁₅P→ ⁰₋₁e+³²₁₆S

write the identity of the missing nucleus for the following nuclear decay reaction:?→84be+01e

?→⁸₄Be+⁰₁e

let us balance the above equation to see

 let me start by rewriting the above equation

^a_zX →⁸₄Be+⁰₁e

a=8+0

a= 8

and z= 4+1

z=5

 the original nuclide have an atomic number  5 an a mass number  8, the atom hat fit the above description is boron-8, the above decay characterize a positron decay, where a proton decays into a neutron and an electron anitneutrino. The atomic number of the parent nuclide decreases by one while the mass number did not change, this is why the uclide beryllium is formed.

 The above equation can be written as follows

⁸₅B →⁸₄Be+⁰₁e

provide a symbol for the isotope of bismuth that contains 124 neutrons.

 The atom bismuth has an  atomic number of

?→237 93np+42he+00γ

 The above equation can be balanced and we can therefore find the missing atom,

 Let us start by rewriting the  decay equation

?→²³⁷₉₃Np+⁴₂He+ ⁰₀γ

Let us start by rewriting the equation

^a_zX →²³⁷₉₃Np+⁴₂He+ ⁰₀γ

a=237+4+0 =241

and

z =93+2+0=95

 the nuclide that decayed has an atomic number of 5 and a mass number of 241, the nuclide this property is americium-241, the atom decay is two, alpha decay and gamma decay , in alpha decay the atomic number of the parent atom decreases by 2 while the mass number decreases by 4, the nuclide formed will therefore have an atomic number of 95 and a mass number of 241, the atom with the above characteristics is americium-241

 the above decay equation can be rewritten as follows

²⁴¹₉₅Am →²³⁷₉₃Np+⁴₂He+ ⁰₀γ

Solve 5927co+ → 5625mn 42he

 When manganese undergo an alpha decay , the atomic number decreass by 2 while the mass number decreases by 4, the nuclide formed  will therefore have an atomic number of 25 an a mass number of 55, the atom that correpon to this is manganese-55

 Let us write the decay equation

⁵⁹₂₇Co→ ⁵⁶₂₅Mn +⁴₂He

write the identity of the missing nucleus for the following nuclear decay reaction: 224 88ra→42he+?

Let us rewrite the above nuclear equations

²²⁴ ₈₈Ra→⁴₂He+?

LET US BALANCE THE equation

 We will have to rewrite the equation again

²²⁴₈₈Ra→⁴₂He+ ^a_zX

224 = 4+a

A=224-4

A=220

And

For the subscript  we have

88= 2+z

Z=86

 Therefore the nuclide formed will have an atomic number of 86 and a mass number of  220.

 the above equation will therefore be

²²⁴₈₈Ra→⁴₂He+ ²²⁰₈₆Rn

  The mass number of radium decreases by four while the atomic number decreases by 2,  this osa characteristics of an alpha decay.