Gravity increases when:

Consider these ground-state ionization energies of oneelectron species:

SVETLANKA909090 [29]

Answer:

Explanation:

general expression for the ionization energy of any one-electron species.

= z² x ground level energy / n²

z is atomic no , n is energy level .

ground level energy of

H = 1.31 x 10³ kJ/mol

He⁺ = 2² x 1.31 x 10³ kJ/mol = 5.24 x 10³ kJ/mol

Li²⁺ = 3² x 1.31 x 10³ kJ/mol = 1.18 x 10⁴ kJ/mol

b ) the ionization energy of B⁴⁺.

= 5² x 1.31 x 10³ kJ/mol = 32.75 x 10³ kJ/mol

c ) minimum energy required to remove the electron from the n = 3 level of He⁺ per mole

= 5.24 x 10³ / 9 kJ/mol

= 5.82 x 10² kJ/mol

= 5.82 x 10² x 10³ / 6.02 x 10²³ J

.9667 X 10⁻¹⁸ J

= .9667 X 10⁻¹⁸ / 1.6 X 10⁻¹⁹ eV

= 6.042 eV

= 1237.5 / 6.042

= 204.82 nm

=

d )

3 0

3 years ago

Match the following family of elements to its group.

docker41 [41]

Explanation:

To solve this problem, we simply use the periodic table of elements which groups elements based on their atomic numbers.

The atomic number of an element is the number of protons it contains. The protons are the positively charged particles within an atom.

The vertical arrangement of elements on the periodic table is the group.
The horizontal arrangement of elements is the period.

Now;

Noble gases belongs to group 18

Alkali earth metals belongs to group 2

Halogens belongs to group 17

Alkali metals belongs to group 1

Transition metals belongs to group 3-12

7 0

3 years ago

150 mL of 0.25 mol/L magnesium chloride solution and 150 mL of 0.35 mol/L silver nitrate solution are mixed together. After reac

Murljashka [212]

Answer:

$0.175\; \rm mol \cdot L^{-1}$ .

Explanation:

Magnesium chloride and silver nitrate reacts at a $2:1$ ratio:

$\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)$ .

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

$\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}$ .

The precipitate silver chloride $\rm AgCl$ is insoluble in water and barely ionizes. Hence, $\rm AgCl\!$ isn't rewritten as ions.

Net ionic equation:

$\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}$ .

Calculate the initial quantity of nitrate ions in the mixture.

$\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}$ .

Since nitrate ions $\rm {NO_3}^{-}$ do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

$n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol$ .

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be $(1/2)$ of the concentration in the original solution.

$\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}$ .

6 0

3 years ago

he combustion of propane (C3H8) is given by the balanced chemical equation C_3H_8+5O_2\longrightarrow3CO_2+4H_2O C 3 H 8 + 5 O 2

Basile [38]

Answer:

290 grams

Explanation:

Let's begin by writing the balanced chemical equations:

$C_{3}H_{8} +5O_{2} --->3CO_{2} +4H_{2}O$

Then we calculate the number of moles in 97g of propane.

n(propane)= $\frac{mass}{molarmass} =\frac{97g}{44.1g/mol}=2.1995mol$

According to the balanced chemical equation, one mole of propane produces 3 moles of carbon dioxide. So the available number of moles of propane must be multiplied by three to work out the number of carbon dioxide produced.

n(carbon dioxide)= 2.1995mol*3 = 6.5985mol

mass(carbon dioxide) = moles * molar mass

= 6.5985 mol * 44.01 g/mol

= 290 grams

4 0

3 years ago

How many molecules are in 3.50 moles of H2O

Debora [2.8K]

Answer:

2.11 x 10²⁴ molecules.

Explanation:

It is known that every 1.0 mole of a molecule contains Avogadro's number of molecules (NA = 6.022 x 10²³).

Using cross multiplication:

1.0 mole of H₂O contains → 6.022 x 10²³ molecules.

3.5 mole of H₂O contains → ??? molecules.

∴ 3.5 mole of H₂O contain = (3.5 mol)(6.022 x 10²³) = 2.11 x 10²⁴ molecules.

5 0

3 years ago