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3 years ago

Elements can be identified by their

Chemistry

2 answers:

AfilCa [17]3 years ago

7 0

Answer:

physical and chemical properties

Explanation:

Have a great day :D

NARA [144]3 years ago

3 0

Answer:

C) Physical and chemical properties

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A solution contains 15.27 grams of NaCl in 0.670 kg water at 25 °C. What is the vapor pressure of the solution?

Anvisha [2.4K]

Answer:

The vapor pressure of the solution is 23.636 torr

Explanation:

$P_{solution} = X_{solvent}*P_{solvent}$

Where;

$P_{solution$ is the vapor pressure of the solution

$X_{solvent$ is the mole fraction of the solvent

$P_{solvent$ is the vapor pressure of the pure solvent

Thus,

15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl

0.67 kg of water = [(0.67*1000)/(18)]moles = 37.222 moles of H₂O

Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)

Mole fraction of solvent (water) = (37.222)/(37.222+0.261)

Mole fraction of solvent (water) = 0.993

<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm

Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm

the vapor pressure of the solution = 0.0311 atm = 23.636 torr

5 0

3 years ago

Read 2 more answers

Write ionic equations for any three of the following:

worty [1.4K]

The ionic equation is:

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H2O(l)

The required dissociations are

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

So, the ionic equation is

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

The ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

The required dissociations are

HCl(aq) → H⁺(aq) + Cl⁻(aq)

KOH(aq) → K⁺(aq) + OH⁻(aq)

KCl(aq) → K⁺(aq) + Cl⁻(aq)

So, the ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

The ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

H₂SO₄(aq) + Mg(OH)₂(aq) → MgSO₄(aq) + 2H2O(l)

The required dissociations are

H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)

Mg(OH)₂(aq) → Mg²⁺(aq) + 2OH⁻(aq)

MgSO₄(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

So, the ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

Learn more about ionic equations here:

brainly.com/question/11628165

5 0

2 years ago

Is temperature beginning to rise a physical or chemical change?

Lelu [443]

Answer:

If temperature increases, as it does in most reactions, a chemical change is likely to be occurring. This is different from the physical temperature change. During a physical temperature change, one substance, such as water is being heated.

Explanation:

6 0

3 years ago

Read 2 more answers

. The compound Fe(NO3)3(s) is classified as a(n):?. a. polyatomic compound. b. molecular compound. c. ionic compound. d. multi-a

patriot [66]

<span>the best answer is C i.e is ionoic compound. but all other option sare quite close enough but option B is sure wrong. because A molecular compound does not separate in a solvent.</span>

5 0

3 years ago

A process at constant T and P can be described as spontaneous if ΔG < 0 and nonspontaneous if ΔG > 0. Over what range of t

creativ13 [48]

Answer:

Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

2 SO3(g) → 2 SO2(g) + O2(g)

ΔHf° (kJ mol-1) -395.7 -296.8

S° (J K-1 mol-1) 256.8 248.2 205.1

ΔH° = kJ

S° = J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

where ν is the stoichiometric coefficient in the balanced chemical equation.

For ΔS we have likewise

ΔrxnS = ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) = 2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ = 197.8 - T(0.1879)

and the reaction will become spontaneous when the term T(0.1879) becomes greater that 197.8,

0 = 197.8 - 0.1879 T ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

4 0

3 years ago