Any ions with a charge (3+ ) has?

Which metal does not form cations of differing charges?

valkas [14]

Transition metals

Most transition metals differ from the metals of Groups 1, 2, and 13 in that they are capable of forming more than one cation with different ionic charges. As an example, iron commonly forms two different ions

8 0

3 years ago

If the [H3O+] of a solution is 1x 10-8 mol/L the [OH-] is

Studentka2010 [4]

Answer:

(H30+)= 1x10^-6 M

Explanation:

Both pH and pOH have a relationship to belonging to the same aqueous solution: the expression of the Kwater (ionic product of the water Kw) is used:

1x 10-8 mol/L equals to1x10-8 M

(H3O+) x (OH-) = 1x10^-14

(H30+)x 1x 10^-8 =1x10^-14

(H30+)= 1x10^-14/1x 10^-8

(H30+)= 1x10^-6 M

6 0

3 years ago

A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at

lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

Explanation:

Step 1: Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

Step 2: Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

Step 3: Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25 = 8.2 °C

q = 1200 * 4.184 * 8.2 = 41170.56 J

Step 4: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

Step 5: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

Step 6: Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is -5485.7 kJ/mol

8 0

3 years ago

Which elements, when they have to, can have more than

inn [45]

Elements in the third row can break the octet rule

6 0

3 years ago

What is the empirical formula of the compound that is 25.3% magnesium and 74.7% chlorine

Margarita [4]

Answer: $MgCl_2$

Explanation:

25.3% Mg

74.7% Cl

First step: change % to g

25.3g Mg

74.7g Cl

Second step: calculate g/mol of each compound. You can do this by using the atomic mass.

$25.3gMg(\frac{1mol}{24.30g})=1.04mol$

$74.7gCl(\frac{1mol}{35.45g} )=2.11mol$

Third step: determine the lowest number and divide everything by it. Of the result, extract whole number only.

$Mg=\frac{1.04}{1.04} =1$

$Cl=\frac{2.11}{1.04}=2$

Fourth step: Write each compound with their respective number below.

This empirical formula should be: $MgCl_2$

4 0

3 years ago