Answer: option B
Explanation: since nuclear fission involves the decay of larger nuclide into smaller nuclei along with Neutron when it is collide with Neutron.
Example Decay of U-235 into Kr and Ba along with 3 neutrons
Potassium chloride reacts with ammonium nitrate to give ammonium chloride and potassium nitrate.
This is a type of double displacement reaction. The balanced chemical equation can be represented as,
Total ionic equation for this reaction will be,
There is no apparent reaction as this reaction is not accompanied by the formation of a gas or a solid precipitate. We cannot observe any visual reaction as there is not net reaction taking place. All the ions remain as spectator ions.
Answer:
Increasing the pressure on a reaction involving reacting gases increases the rate of reaction. Changing the pressure on a reaction which involves only solids or liquids has no effect on the rate.
Explanation:
Answer:
The object placed in the water has a volume of 19 cm³
Explanation:
<u>Step 1: </u>Data given
volume of the cylinder before adding the object = 28 mL = 28 cm³
After adding an object with volume X the volume rises to 47 mL = 47 cm³
<u>Step 2:</u> Calculate the volume of the object
Volume of the object = Final volume - initial volume
Volume of the object = 47 cm³ (or 47 mL) - 28 cm³ ( or 28 mL) = 19 cm³ (or 19 mL)
The object placed in the water has a volume of 19 cm³
Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
i hope that this eg gonna help u
