All categories

2 years ago

Why does fluorine have a higher ionization energy than iodine?

1 answer:

7 0

Fluorine has a higher ionization energy because it is in a lower energy level and does not receive the same intensity of shielding as Iodine. Thus Fluorine has a stronger nuclear force on the surrounding electrons and require more energy to remove electrons.

You might be interested in

Which two forces keep the planets and moons in their orbits?

jeyben [28]

Answer:

“Without gravity, the planets' inertia would keep them moving in straight lines. Gravity “steers” the planets in their oval paths around the Sun. Together, gravity and inertia keep the planets in their orbits.”

Explanation:

BRAINLIEST PLZZZ

5 0

2 years ago

Read 2 more answers

The combustion of 0.590 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

Alenkinab [10]

Answer:

2.943 °C temperature change from the combustion of the glucose has been taken place.

Explanation:

Heat released on combustion of Benzoic acid; :

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.590 g

Moles of benzoic acid = $\frac{0.590 g}{122.12 g/mol}=0.004831 mol$

Energy released by 0.004831 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004831 mol=15.5955 kJ=15,595.5 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.125°C

$Q=C\times \Delta T$

$15,595.5 J=C\times 2.125^oC$

$C=7,339.05 J/^oC$

Heat released on combustion of Glucose: :

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=1.400 g

Moles of glucose = $\frac{1.400 g}{180.16 g/mol}=0.007771 mol$

Energy released by the 0.007771 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.007771 mol=21.6030 kJ=21,603.01 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$21,603.01 J=7,339.05 J/^oC\times \Delta T'$

$\Delta T'=2.943^oC$

2.943 °C temperature change from the combustion of the glucose has been taken place.

8 0

3 years ago

Consider the reaction mg(oh)2(s)→mgo(s)+h2o(l) with enthalpy of reaction δhrxn∘=37.5kj/mol what is the enthalpy of formation of

Anni [7]

Answer is: -601,2 kJ/mol
Chemical reaction: Mg(OH)₂ → MgO + H₂O.
ΔHrxn = 37,5 kJ/mol.
ΔHf(Mg(OH)₂) = −924,5 kJ/mol.
ΔHf(H₂O) = −285,8 kJ/mol.
ΔHrxn -enthalpy of reaction.
ΔHf - enthalpy of formation.
ΔHrxn=∑productsΔHf−∑reactantsΔHf.
ΔHf(MgO) = -924,5 kJ/mol - (-285,8 kJ/mol) + 37,5 kj/mol.
ΔHf(MgO) = -601,2 kJ/mol.

7 0

3 years ago

Read 2 more answers

For this exercise, you can simulate the described conditions by changing the values in the run experiment tool of the simulation

Furkat [3]

1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.

3 0

2 years ago

Read 2 more answers

Assume that a daily diet of 2000 calories (i.e. 8.37 x 106 J) is converted completely to body heat.

slava [35]

Answer:

(a) the mass of the water is 3704 g

(b) the mass of the water is 199, 285.7 g

Explanation:

Given;

Quantity of heat, H= 8.37 x 10⁶ J

Part (a) mass of water (as sweat) need to evaporate to cool that person off

Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg

H = m x Lvap.

$m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg$

mass in gram ⇒ 3.704 kg x 1000g = 3704 g

Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J

specific heat capacity of water, C, 4200 J/kg.°C

H = mcΔθ

where;

Δθ is the change in temperature = 35 - 25 = 10°C

$m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg$

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g

5 0

3 years ago