What's the force between two moving objects that are touching called

The names and chemical formulae of some chemical compounds are written in the first two columns of the table below. Each compoun

charle [14.2K]

Answer:

See below

Explanation:

Name Formula Major species

Zinc iodide ZnI₂ H₂O(ℓ), I⁻(aq), Zn²⁺(aq),

Nitrogen(I) oxide N₂O H₂O(ℓ), N₂O(aq)

Sodium nitrite NaNO₂ H₂O(ℓ), Na⁺(aq), NO₂⁻(aq)

Glucose C₆H₁₂O₆ H₂O(ℓ), C₆H₁₂O₆(aq)

Nickel(II) iodide NiI₂ H₂O(ℓ), I⁻(aq), Ni²⁺(aq)

Glucose and nitrogen(I) oxide are covalent compounds. They do not dissociate in solution.
The compounds containing metals are ionic. They produce ions in solution.
ZnI₂ and NiI₂ produce twice as many iodide ions as metal ions.

6 0

3 years ago

How many parts per million of fluoride in a solution that is 500 grams of fluoride and 500,000 liters water

erik [133]

0.000001‬ppm

Explanation:

Mass of fluoride = 500g

Volume of water = 500000liters

Unknown:

Parts per million of fluoride = ?

Solution:

The parts per million is the amount of solute in milligram dissolved in a liter of water or milligram per kilogram of solvent

It is a unit used to express very small concentration.

we need to convert g - mg

500g = 500 x 10⁻³mg = 0.5mg

Concentration in parts per million = $\frac{mass in mg}{Volume in liters}$

Concentration in parts per million = $\frac{0.5}{500000}$ = 0.000001‬ppm

learn more:

Parts per million brainly.com/question/2854033

#learnwithBrainly

7 0

3 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

How many liters of oxygen are in 8.32 moles of oxygen at STP?

Lana71 [14]

Answer:

for more details are in the pic

4 0

3 years ago

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose

Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution = 101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0

3 years ago