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Tresset [83]
4 years ago
7

Nuclear fission occurs when an atomic nucleus splits into two or more smaller and less massive atomic nuclei.

Chemistry
2 answers:
Jobisdone [24]4 years ago
8 0
I beleive the answer is true
ololo11 [35]4 years ago
8 0
I think The answer is true
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What cannot be broken down into other substances?
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Answer:

element

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How many grams are in 0.50 moles of water
adell [148]

Answer:

9.01 grams

Explanation:

~18.02 grams are in 1 mole of water so you'd divide that in half to get 9.01 grams.

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2 years ago
A simple machine produces 25 joules of output work for every 50 joules of input work . What is the efficiency of this machine ?
nataly862011 [7]
2:1, output is half the input
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4 years ago
Equation below is used to determine the heat flux for convection
OLEGan [10]
Q=hc A (Ts-Ta)
q = heat transfered per unit time
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4 years ago
If all of the energy from burning 281.0 g of propane (ΔHcomb,C3H8 = –2220 kJ/mol) is used to heat water, how many liters of wate
lapo4ka [179]

This problem is providing us with the mass of propane, its enthalpy of combustion, and the initial and final temperature of water that can be heated from the burning of this fuel. At the end, the result turns out to be 42.27 L.

<h3>Combustion:</h3>

In chemistry, combustion reactions are based on the burning of fuels by using oxygen and producing both carbon dioxide and water. For propane, we will have:

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

Hence, we can calculate the heat released from this reaction by using the mass, which has to be converted to moles, and the given enthalpy of combustion:

Q=281.0g*\frac{1mol}{44.09g}*-2220\frac{kJ}{mol}*\frac{1000J}{1kJ}\\ \\ Q=-1.415x10^7 J

<h3>Calorimetry:</h3>

In chemistry, we can analyze the mass-specific heat-temperature-heat relationship via the most general heat equation:

Q=mC\Delta T

Thus, since Q was obtained from the previous problem, but the sign change because the released heat is now absorbed by the water, one can calculate the mass of water that rises from 20.0°C to 100.0°C with this heat:

m=\frac{Q}{C\Delta T} =\frac{1.415x10^7J}{4.184\frac{J}{g\°C}(100.0\°C-20.0\°C)}\\ \\m=4.227x10^4g

Finally, we convert it to liters as required:

V=4.227 x10^4g*\frac{1mL}{1.00g}*\frac{1L}{1000mL}  \\\\V=42.27L

Learn more about calorimetry: brainly.com/question/1407669

4 0
2 years ago
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