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nika2105 [10]
2 years ago
12

Which characteristic do electronegativity differences indicate about reactions between atoms

Chemistry
1 answer:
emmasim [6.3K]2 years ago
6 0
O valence electron number is the answer
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Suppose that 100 grams of water at 50.0°C is placed in contact with 200 grams of iron at 30.0°C. The final
wel

Answer:

The answer would be 1.5 kJ.

Explanation:

When you use the equation q = m x c x ∆T you will be able to find the energy gained or lost. The data for the water in this case is just there to distract you so ignore it. :D

4 0
2 years ago
Why is it important to analyze data?
gayaneshka [121]

Answer:

It is important to analyze data to further understand what's going on. By analyzing data, you know more about what you are investigating. Without analyzing data, you may find it harder to figure something.

7 0
3 years ago
A machine
rosijanka [135]

i believe the answer is A (((:

4 0
2 years ago
Read 2 more answers
Write the molecular formula for the following compound.<br>​
Sholpan [36]

Answer : The molecular formula of the compound is, C_8H_9O_3N

Explanation :

Molecular formula : It is the representation of substance by the symbols and it denotes the number of atoms of each element present in the compound.

Now count the number of carbon, hydrogen, nitrogen and oxygen atoms present in the given compound.

As we see that in the given compound, there are 8 atoms of carbon element, 3 atoms of oxygen element, 1 atom of nitrogen element, 9 atoms of hydrogen element.

Thus, the molecular formula of the compound will be C_8H_9O_3N

7 0
3 years ago
A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume
pshichka [43]

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure   = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol

amount of SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol

mole fraction of N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330

mole of fractionSF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669

PV = nRT

P of N_2 O = \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa

mole of SF_4=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa

Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

6 0
3 years ago
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