What is water molecules (H2O) an example of?

Solid particles are moving (slower or faster) than liquid and thus have (more or less) energy

Naddik [55]

Answer: faster than , less energy

4 0

4 years ago

Give the name and formula of the compound formed from the following elements: a) Sodium and nitrogen b) Oxygen and strontium c)

elena-s [515]

Answer: a) Sodium and nitrogen : $NaN_3$ : sodium nitride.

b) Oxygen and strontium : $SrO$ : strontium oxide.

c) Aluminum and chlorine : $AlCl_3$ : aluminium chloride.

d) Cesium and bromine : $CsBr$ : cesium bromide.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

a) Sodium and nitrogen : Here sodium is having an oxidation state of +1 called as $Na^{+}$ cation and nitrogen forms an anion $N^{3-}$ with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $NaN_3$ named as sodium nitride.

b) Oxygen and strontium : Here strontium is having an oxidation state of +2 called as $Sr^{+}$ cation and oxygen forms an anion $O^{2-}$ with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $SrO$ named as strontium oxide.

c) Aluminum and chlorine : Here aluminium is having an oxidation state of +3 called as $Al^{3+}$ cation and chlorine forms an anion $Cl^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $AlCl_3$ named as aluminium chloride.

d) Cesium and bromine : Here cesium is having an oxidation state of +1 called as $Cs^{+}$ cation and bromine forms an anion $Br^{-}$ with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $CsBr$ named as cesium bromide.

7 0

3 years ago

When the free-energy change (AG) of a reaction is negative, the reaction is said to be: Endergonic Non spontaneous Spontaneous E

Greeley [361]

Answer:

Spontaneous

Explanation:

For a spontaneous reaction , the value for the change in free energy or gibbs' free energy is negative .

The change in free energy , i.e. , ΔG , denotes the maximum amount of usable energy released , as going from initial state , i.e. , the reactant towards the final state , i.e. , the product .

And , the sign of the ΔG , determines whether the reaction is Spontaneous or non Spontaneous or at equilibrium ,

i.e. ,

if

ΔG < 0 , the reaction is Spontaneous

ΔG > 0 , the reaction is non Spontaneous

ΔG = 0 , the reaction is at equilibrium

For ,

ΔG < 0 , that the reaction proceed without any energy input , hence , it is Spontaneous in nature .

6 0

3 years ago

How many grams are in 5.00 * 10^25 atoms of helium

statuscvo [17]

Answer:

dfghgvfdsrftgbnbvcdfvgbhnbvcxsdefrgthbvcxdsfgt

Explanation:

sdfgvfcdsertgyhbvcfxdsrftgh

4 0

3 years ago

A 27.86 ml sample of 0.1744 m hno3 is tirades with 29.4ml of a job solution. What is the molarity of the koh

Arte-miy333 [17]

Answer:

0.165 mol·L⁻¹

Explanation:

1. Write the chemical equation for the reaction.

HNO₃ + KOH ⟶ KNO₃ + H₂O

===============

2. Calculate the moles of HNO₃

c = n/V Multiply each side by V and transpose

n = Vc

V = 0.027 86 L

c = 0.1744 mol·L⁻¹ Calculate the moles of HNO₃

Moles of HNO₃ = 0.027 86 × 0.1744

Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃

===============

3. Calculate the moles of KOH

1 mol KOH ≡ 1 mol HNO₃ Calculate the moles of KOH

Moles of KOH = 4.859 × 10⁻³× 1/1

Moles of KOH = 4.859 × 10⁻³ mol KOH

===============

4. Calculate the molar concentration of the KOH

V = 29.4 mL = 0.0294 L Calculate the concentration

c = 4.859 × 10⁻³/0.0294

c = 0.165 mol·L⁻¹

4 0

4 years ago