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miv72 [106K]
3 years ago
11

If one mole (1.00 mol) of copper (Cu) has 6.02x10^23 atoms of copper, how many molecules are in 1.00 mol of nitrogen gas (N2)

Chemistry
1 answer:
sveta [45]3 years ago
5 0
WAS IT C??? CANT BE FOR SURE

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How are catapults an example of a third class lever? Please explain!
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How does reproduction leads to natural selection in a population
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2 years ago
When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro
Bas_tet [7]
<span>Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
7 0
3 years ago
Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in
AysviL [449]

Answer:

Amount of heat absorbed by water is 2604.54 J.

Explanation:

Amount of heat absorbed by water = m_{water}\times C_{water}\times \Delta T_{water}

where m represents mass, C represents specific heat and \Delta T represents change in temperature.

Here m_{water}=75.0 g , C_{water}=4.184J/(g.^{0}\textrm{C}) and \Delta T = (final temperature - initial temperature) = (29.5-21.2) ^{0}\textrm{C} = 8.3 ^{0}\textrm{C}

So, amount of heat heat absorbed by water

     = (75.0g)\times (4.184\frac{J}{g.^{0}\textrm{C}})\times (8.3^{0}\textrm{C})

     = 2604.54 J

8 0
3 years ago
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