multiply the number of each atom with its molecular mass. (see the periodic table)
5*12 + 12*1 = 72 gram per mole.
Or more accurately: 5*12.01078 + 12*1.007947 = 72.149264 g/mole
Answer:
323.15 °C
Explanation:
Considering the ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.
P ∝ T
Also,
Using Charle's law
Given ,
P₂ = 2P₁
T₁ = 25 °C
T₂ = ?
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (25 + 273.15) K = 298.15 K
Using above equation as:
New temperature = 596.3 K
Also,
T(K) - 273.15 = T( °C)
<u>So, Temperature = 596.3 - 273.15 °C = 323.15 °C</u>
Answer:
2.4506 x 
Explanation:
move decimal point to the right 1 time : 24506
24506 becomes 2.4506
count how many numbers after decimal point and that becomes b
b = 4
0.019moles
Explanation:
Given parameters:
Mass of Fe₂O₃ = 1.5g
Unknown:
Number of moles of Fe produced = ?
Solution:
To solve this problem, we must work from the known to the unknown. The known specie here is the mass of Fe₂O₃.
First inspect the equation and balance it;
2Fe₂O₃ + 3C → 4Fe + 3CO₂
Now find the number of moles of Fe₂O₃
Number of moles of Fe₂O₃ = 
molar mass of Fe₂O₃ = 2(56) + 3(16) = 112 + 48 = 160g/mol
Number of moles = 
= 0.009375moles
From the equation, we see that;
2 moles of Fe₂O₃ produced 4 moles of Fe
0.009375 moles of Fe₂O₃ will produce 
= 0.01875moles of Fe
To 2 decimal places;
0.019moles
learn more:
Number of moles brainly.com/question/1841136
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Answer:In alpha decay, shown in Fig. 3-3, the nucleus emits a 4He nucleus, an alpha particle. Alpha decay occurs most often in massive nuclei that have too large a proton to neutron ratio. An alpha particle, with its two protons and two neutrons, is a very stable configuration of particles. Alpha radiation reduces the ratio of protons to neutrons in the parent nucleus, bringing it to a more stable configuration. Many nuclei more massive than lead decay by this method.
Consider the example of 210Po decaying by the emission of an alpha particle. The reaction can be written 210Po Æ 206Pb + 4He. This polonium nucleus has 84 protons and 126 neutrons. The ratio of protons to neutrons is Z/N = 84/126, or 0.667. A 206Pb nucleus has 82 protons and 124 neutrons, which gives a ratio of 82/124, or 0.661. This small change in the Z/N ratio is enough to put the nucleus into a more stable state, and as shown in Fig. 3-4, brings the "daughter" nucleus (decay product) into the region of stable nuclei in the Chart of the Nuclides.
In alpha decay, the atomic number changes, so the original (or parent) atoms and the decay-product (or daughter) atoms are different elements and therefore have different chemical properties.
Upper end of the Chart of the Nuclides
In the alpha decay of a nucleus, the change in binding energy appears as the kinetic energy of the alpha particle and the daughter nucleus. Because this energy must be shared between these two particles, and because the alpha particle and daughter nucleus must have equal and opposite momenta, the emitted alpha particle and recoiling nucleus will each have a well-defined energy after the decay. Because of its smaller mass, most of the kinetic energy goes to the alpha particle.
