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3 years ago

How does the idea of a neutron help address the puzzles about the protons in the nucleus and the mass of atoms?

Chemistry

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

hen the number of neutrons is known and the atomic number of an element is known, it becomes easier to determine the approximate mass number by adding the two.

Explanation:

Hope it shelps

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Is there a phobia for chemical reactions specifically?

gizmo_the_mogwai [7]

Answer:

chemonoia

Explanation:

Many people have a strong fear of “chemicals” that belies the evidence. It's called chemonoia – and it may be damaging their health. If you were paranoid, you might think there really was something in the water that's damaged our sense of reason.

8 0

3 years ago

Which is the least dense?

AVprozaik [17]

1000 g of feathers is the least dense

6 0

3 years ago

Which of the following is an example of a concentric contraction

anastassius [24]

The best and the correct answer among the choices provided by the question is the second choice or letter B.

A bicep curl is a<span>n example of a concentric contraction.

</span>I hope my answer has come to your help. Have a nice day ahead and may God bless you always!

3 0

3 years ago

Read 2 more answers

A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of

Stella [2.4K]

Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

$n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC$

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

$n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH$

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

$m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH} =0.1900gO$

4. Compute the moles of oxygen by using its molar mass:

$n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO$

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

$C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1$

6. Search for the closest whole number (in this case multiply by 2):

$C_3H_6O_2$

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

$M=12*3+1*6+16*2=74g/mol$

Which is about three times in the molecular formula, for that reason, the actual formula is:

$C_9H_{18}O_6$

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

6 0

3 years ago

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

Answer : 7.87 X $10^{-6}$ .

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, $\frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )$

on solving with the given values , $Q_{s}= 1.86 eV$ and T as 645 + 273 K and rest are the constants.

$\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )$

we get the answer as 7.87 X $10^{-6}$ .

6 0

3 years ago

Read 2 more answers