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olga55 [171]
3 years ago
6

What happens to the mass of atoms during a nuclear reaction.

Chemistry
1 answer:
Damm [24]3 years ago
5 0


energy can't be created or destroyed, but mass particles can decay or tranform into others as long as the total mass/energy (remeber E=MC2) is conserved, in nuclear reactions one type of sub-atomic particle change into another, this destabilizes the nucleus forcing it to split and release a certain amount of energy in the form of massless particles
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Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h
cluponka [151]
<span>C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5</span>
3 0
3 years ago
You have dissolved 10 g sodium oxide in 200 ml water.calculate concentration of the solution
e-lub [12.9K]

Answer:

0.85 Molar Na2O

Explanation:

Determine the moles of sodium oxide, Na2O, in 10 grams by dividing by the molar mass of Na2O (61.98 g/mole).

 (10 g Na2O)/(61.98 g/mole) = 0.161 moles Na2O.

Molar is a measure of concentration.  It is defined as moles/liter.  A 1 M  solution contains 1 mole of solute per liter of solvent.  [200 ml water = 0.2 Liters water.]

In this case, we have 0.161 moles Na2O in 0.200 L of solvent.

 (0.161 moles Na2O)/(0.200 L) = 0.85 Molar Na2O

8 0
2 years ago
How many molecules are in 2.00 moles of H20????????
Elis [28]

2x 6.022x10^23= 1.204x10^24

7 0
3 years ago
How is the structure of atoms altered during fission and fusion?
Dmitriy789 [7]
At first sight it doesn't bode well. The key is in how firmly the protons and neutrons are held together. In the event that an atomic response produces cores that are more firmly bound than the firsts then vitality will be created, if not you should place vitality into make the response happen.
6 0
3 years ago
Cuando se quema 1 mol de metano –o sea, 16 g–, se desprenden 802
Anvisha [2.4K]

Answer:

1 gramo de metano aporta 50.125 kilojoules.

1 gramo de metano aporta 48.246 kilojoules.

Explanation:

La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (Q), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (\bar {Q}), en kilojoules por mol, dividido por su masa molar (M), en gramos por mol:

Q = \frac{\bar Q}{M} (1)

A continuación, analizamos cada caso:

Metano

Q = \frac{802\,\frac{kJ}{mol} }{16\,\frac{g}{mol} }

Q = 50.125\,\frac{kJ}{g}

1 gramo de metano aporta 50.125 kilojoules.

Octano

Q = \frac{5500\,\frac{kJ}{mol} }{114\,\frac{g}{mol} }

Q = 48.246\,\frac{kJ}{mol}

1 gramo de metano aporta 48.246 kilojoules.

3 0
2 years ago
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