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guajiro [1.7K]
4 years ago
6

The table lists some compounds found in foods and their formulas.

Chemistry
1 answer:
andriy [413]4 years ago
4 0

Answer:

A

Explanation:

the number of molecule won't be twice as fructose has 12 hydrogen molecules and lactose has 22 hydrogen molecules

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In general, ionization energies increase across a period from left to right. Explain why the second ionization energy of Cr is h
rodikova [14]

Answer:So this leads to the fact that second ionization energy  of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d  orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

Explanation:

Ionization energy is the amount of energy that we require to remove an electron form an isolated gaseous atom.

As we move from left to right across a period electrons are added to the same outermost shell therefore the attraction between the electrons and nucleus increases since more number of negatively charged electron are attracted to the positively charged nucleus.  This attraction leads to the decrease in atomic radii across a period and increase in ionization energy .

The increase in ionization energy occurs due to the fact that as the attraction  between the nucleus and outermost electrons increases so the electrons are more tightly bound to the nucleus hence more amount of energy is required to ionize the electron which leads to increase in ionization energy.

The electronic configuration of Cr and Mn are:

Cr:[Ar]3d⁵4S¹

Mn:[Ar]3d⁵4S²

The electronic configuration of Cr and Mn after 1st ionization:

Cr:[Ar]3d⁵4S⁰

Mn:[Ar]3d⁵4S¹

The electronic configuration of Cr and Mn after 2nd ionization:

Cr:[Ar]3d⁴4S⁰

Mn:[Ar]3d⁵4S⁰

As we can see that that 3d orbital of Cr (Chromium) is half filled with 5 electrons in it  and 4s orbital of Cr is also half-filled.

So when Cr is ionized for the first time then the electron from the half-filled 4s orbital will be removed .As the 1 electron present in outer most 4s orbital is removed so the 4s orbital now is completely vacant.

Now for the second ionization energy an electron ahs to be removed from half-filled 3d⁵ orbital. Hunds rule of maximum multiplicity states that the fully-filled or half-filled orbitals have maximum stability on account of symmetry and exchange energy.

So half-filled 3d⁵ orbital of Cr is very stable and hence to remove an electron from this would be require a lot of energy and hence the second ionization energy of chromium is higher than that of Manganese.

In case of Mn  the 3d orbital is also half -filled as chromium but the 4s orbital contains two electrons. when we remove the first electron from this orbital then also there is 1 electron present in the 4s orbital . So for the second ionization of Mn the only electron left in 4s orbital will be removed as the removal of electron from a 4s orbital is much easier as it requires less amount of energy as compared to  removal of  a electron from stable half filled 3d orbital.

So this leads to the fact that second ionization energy  of chromium is higher as compared to that of Manganese because of the unavailability of electron in the outermost orbital in case of chromium so the second electron has to be removed form the stable half filled 3d  orbital which requires more energy. Whereas in case of Manganese there is an electron available in outermost 4s orbital.

3 0
3 years ago
Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volum
zhuklara [117]

Answer:

The volume of nitrogen oxide formed is 35.6L

Explanation:

The reaction of nitric acid with copper is:

Cu(s) + 4HNO₃ → Cu(NO₃)₂ + 2NO₂(g) + 2H₂O(l)

Moles of copper are:

4.95cm^3\frac{8.95g}{1cm^3} \frac{1mol}{63.55g} = 0.697 moles

Moles of nitric acid are:

230mL\frac{1.42g}{mL} \frac{68g}{100g} \frac{1mol}{63.01g}=3.52moles

As 1 mol of Cu reacts with 4 moles of HNO₃:

0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.

Moles of NO₂ produced are:

0.697 mol Cu × (2mol NO₂ / 1mol Cu) = <em>1.394 moles of NO₂</em>

Using PV = nRT

<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>

Thus, volume is:

V = nRT / P

V = 1.394mol×0.082atmL/molK×301.35K / 0.967atm

V = 35.6L

<em>The volume of nitrogen oxide formed is 35.6L</em>

3 0
3 years ago
7. If one container holds 3.45 gallons of gas, how many containers would I need to hold 17
NNADVOKAT [17]

Answer:

5

Explanation:

if you do 17/3.45

=4.9 but if you round it up to the nearest whole number 5.

3.45*5=17.25

it equals to 17.25

7 0
3 years ago
A 45.0 g block of an unknown metal is heated in a hot water bath to 100.0°c. when the block is placed in an insulated vessel con
professor190 [17]
Let's apply the principle of conservation of energy.

Heat of metal = Heat of water
mCmetalΔT = mCwaterΔT

Applying the given values,

(45 g)(Cmetal)(100 - 28 °C) = (130 g)(4.18 J/g-°C)(28 - 25 °C)
Solving for Cmetal,
<em>Cmetal = 0.503 J/g°C

Therefore, the heat capacity of the unknown metal is 0.503 J/g</em>
°C.
5 0
4 years ago
Which of the following elements is the largest? :
Troyanec [42]

Answer: oxygen

Explanation: as you go across period on the periodic table, the elements get larger. Their atomic masses also increase meaning it gets heavier and thus larger.

7 0
3 years ago
Read 2 more answers
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