After 25 days, it remains radon 5.9x10^5 atoms.
Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.
N(Ra) = 5.7×10^7; initial number of radon atoms
t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days
n = 25 days / 3.8 days
n = 6.58; number of half-lifes of radon
N1(Ra) = N(Ra) x (1/2)^n
N1(Ra) = 5.7×10^7 x (1/2)^6.58
N1(Ra) = 5.9x10^5; number of radon atoms after 25 days
The half-life is independent of initial concentration (size of the sample).
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Answer:
25.35%
Explanation:
Again let me restate the the equation of the reaction;
H2O (ℓ) + 2 MnO4 - (aq) + 3 CN- (aq) → 2 MnO2 (s) + 3 CNO- (aq) + 2 OH- (aq)
Amount of potassium permanganate reacted = 10.2/1000 * 0.08035 = 8.1957 * 10^-4 moles
If 2 moles of MnO4 - reacts with 3 moles of CN-
8.1957 * 10^-4 moles of MnO4 - reacts with 8.1957 * 10^-4 * 3/2
= 1.229 * 10^-3 moles of CN-
Mass of CN- reacted = 1.229 * 10^-3 moles of CN- * 26.02 g/mol
= 0.03 g
Hence, percentage of the cyanide = 0.03 g/0.1183 g * 100
= 25.35%
Answer:
Weathering and erosion
Explanation:
Weathering can be explained as the breaking down of rocks/minerals on the surface of the Earth as a result of contact with biological organism, water, air and other factors
. There are 3 common types of weathering which are;
1) physical weathering
2) biological weathering
3) chemical weathering
Erosion can be regarded as a geological process, whereby earthen material are been transported away by natural forces, these forces could be wind as well as water.
Therefore, as you were climbing a a mountain, you noticed that rocks were crumbling below your feet and moving down the mountain. What is observed are weathering and erosion processes.
After the weakening and broken up of the rock by weathering then erosion transport the bit of the rock down the mountain as you are climbing, which means the "weathering process" breakdown and the "erosion process" involves the transport or movement of the bit of the rocks
Answer:
D = 28.2g
Explanation:
Initial temperature of metal (T1) = 155°C
Initial Temperature of calorimeter (T2) = 18.7°C
Final temperature of solution (T3) = 26.4°C
Specific heat capacity of water (C2) = 4.184J/g°C
Specific heat capacity of metal (C1) = 0.444J/g°C
Volume of water = 50.0mL
Assuming no heat loss
Heat energy lost by metal = heat energy gain by water + calorimeter
Heat energy (Q) = MC∇T
M = mass
C = specific heat capacity
∇T = change in temperature
Mass of metal = M1
Mass of water = M2
Density = mass / volume
Mass = density * volume
Density of water = 1g/mL
Mass(M2) = 1 * 50
Mass = 50g
Heat loss by the metal = heat gain by water + calorimeter
M1C1(T1 - T3) = M2C2(T3 - T2)
M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)
0.444M1 * 128.6 = 209.2 * 7.7
57.0984M1 = 1610.84
M1 = 1610.84 / 57.0984
M1 = 28.21g
The mass of the metal is 28.21g
