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2 years ago

As you move from left to right across

Chemistry

1 answer:

Slav-nsk [51]2 years ago

3 0

Answer:

As we move from left to right across the periodic table the number of valance electrons in an atom increase.

Explanation:

As we move from left to right across the periodic table the number of valance electrons in an atom increase.The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

atmospheric pressure at elevations of 8000 feet averages about 0.72 atmospheres. Would a cabin pressurized at 500 mm hg meet fed

NeTakaya

Answer:

Yes, but it must be kept at that value and do not let it to decrease more.

Explanation:

Hello.

In this case, in order to substantiate whether the cabin meet the federal standards, we need to convert the 500 mmHg to atm and compare the result with 0.72 atm by knowing that 1 atm equals 760 mmHg:

$500mmHg*\frac{1atm}{760mmHg} \\\\=0.66atm$

Thus, since 0.66 atm is 0.06 atm away from the federal standard we can infer that it may meet the federal standard, however, it would not be recommended to let the pressure decrease more than that.

8 0

2 years ago

The Goodyear blimp contains 5.7 x 10^6 L of helium at 25 degrees Celsius and 1 atm. What is the mass in grams of the helium insi

anygoal [31]

Answer:

1.72x10⁻⁵ g

Explanation:

To solve this problem we use the PV=nRT equation, where:

P = 1 atm

V = 5.7x10⁶ L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 25 °C ⇒ (25+273.16) = 298.16 K

And we solve for n:

1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

n = 4.29x10⁻⁶ mol

Finally we convert moles of helium to grams, using its molar mass:

4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g

8 0

2 years ago

All elements can be classified as metals, metalloids, or ?

svlad2 [7]

All elements can be classified as metals, metalloids, or nonmetals

5 0

3 years ago

Question 8 of 10

MAVERICK [17]

Answer:

Explanation:

if you start with 12 atoms of carbon and its called the half life of carbon what is half of 12

8 0

2 years ago

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