Answer:
23430.4 J.
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 70 g
Initial temperature (T₁) = 10 °C
Final temperature (T₂) = 90 °C
Specific heat capacity (C) = 4.184 J/gºC
Heat (Q) required =?
Next, we shall determine the change in the temperature of water. This can be obtained as follow:
Initial temperature (T₁) = 10 °C
Final temperature (T₂) = 90 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 90 – 10
ΔT = 80 °C
Finally, we shall determine the heat energy required to heat up the water. This can be obtained as follow:
Mass (M) = 70 g
Change in temperature (ΔT) = 80 °C
Specific heat capacity (C) = 4.184 J/gºC
Heat (Q) required =?
Q = MCΔT
Q = 70 × 4.184 × 80
Q = 23430.4 J
Therefore, 23430.4 J of heat energy is required to heat up the water.
Alpha particles are released by high mass, proton rich unstable nuclei. The alpha particle is a helium nucleus; it consists of two protons and two neutrons. It contains no electrons to balance the two positively charged protons. Alpha particles are therefore positively charged particles moving at high speeds.
Answer:
A. Elements
Elements are substances that cannot be separated into simpler substances. Salt is made up of the elements sodium and chloride. Water is made up of the elements hydrogen and oxygen, to list a few.
The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
<h3>Determine the freezing point of the solution </h3>
First step : Calculate the molality of NaCl
molality = ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )
= 0.186 mol/kg
Next step : Calculate freezing point depression temperature
T = 2 * 0.186 * kf
where : kf = 1.86°c.kg/mole
Hence; T = 2 * 0.186 * 1.86 = 0.69°C
Freezing point of the solution
Freezing temperature of solvent - freezing point depression temperature
0°C - 0.69°C = - 0.69°C
Hence the Freezing temperature of the solution is - 0.69°C
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