Answer:
It is
D- <u><em>DEFINE</em></u> the problem
R-<em><u>RESEARCH</u></em> on the problem
H- Carry out a <u><em>HYPOTHESIS</em></u>
E- carry out an <em><u>EXPERIMENT</u></em>.
R-Analyse the <u><em>RESULT</em></u>
C-summarise the <u><em>CONCLUSION</em></u>.
Explanation:
Hope it helps.
Its a formula relating to specific heat capacity
Δθ refers to the change in temperature
Q refers to the energy neededto raise the temperature of an object by the change in temperature
m stands for the mass of tje object
c is the specific heat capacity which is the amount of energy needed to heat up an object per unit mass
The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
Answer:
44.8 L
Explanation:
Using the ideal gas law equation:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
At Standard temperature and pressure (STP);
P = 1 atm
T = 273K
Hence, when n = 2moles, the volume of the gas is:
Using PV = nRT
1 × V = 2 × 0.0821 × 273
V = 44.83
V = 44.8 L
Answer:
Military
Explanation:
I may be wrong but military seems most likely
