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4 years ago

Calculate the heat needed to be removed to freeze 45g of water at 0C

Chemistry

1 answer:

alexira [117]4 years ago

4 0

Answer: The heat needed to be removed to freeze 45.0 g of water at 0.0 °C is 15.01 KJ.

Explanation:

Firstly, we need to define the term "latent heat" which is the amount of energy required "absorbed or removed" to change the phase "physical state; solid, liquid and vapor" without changing the temperature.
Types of latent heat: depends on the phases that the change occur between them;

Liquid → vapor, latent heat of vaporization and energy is absorbed.
Vapor → liquid, latent heat of liquification and the energy is removed.
Liquid → solid, latent heat of solidification and the energy is removed.
Solid → liquid, latent heat of fusion and the energy is absorbed.

In our problem, we deals with latent heat of freezing "solidification" of water.
The latent heat of freezing of water, ΔHf, = 333.55 J/g; which means that the energy required to be removed to convert 1.0 g of water from liquid to solid "freezing" is 333.55 g at 0.0 °C.
Then the amount of energy needed to be removed to freeze 45.0 g of water at 0.0 °C is (ΔHf x no. of grams of water) = (333.55 J/g)(45.0 g) = 15009.75 J = 15.01 KJ.

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To what are wax susceptible?

inna [77]

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4 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago

How to separate crystals from solution

tatiyna

Answer:

If a saturated hot solution is allowed to cool, the solute is no longer soluble in the solvent and forms crystals of pure compound. Impurities are excluded from the growing crystals and the pure solid crystals can be separated from the dissolved impurities by filtration.

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2 years ago

If D+2 would react with E-", what do you predict to be the formula?

GuDViN [60]

Answer:DE2

Explanation:

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3 years ago

In the reaction Zn H 2SO 4→ ZNSO4 + H2, which, if any element is oxidized

Sedbober [7]

Answer:

I hope this answer is correct

Explanation:

This is an oxidation-reduction (redox) reaction:

Zn0 - 2 e- → ZnII (oxidation)

2 HI + 2 e- → 2 H0 (reduction)

Zn is a reducing agent, H2SO4 is an oxidizing agent.

Reactants:

Names: Zinc source: wikidata, accessed: 2019-09-07, Zinc powder (pyrophoric) source: ICSC, accessed: 2019-09-04, Zn source: wikidata, accessed: 2019-09-07

Appearance: Grey-to-blue powder source: ICSC, accessed: 2019-09-04

H2SO4 – Sulfuric acid source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-07source: NIOSH NPG, accessed: 2019-09-02

Other names: Oil of vitriol source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-07source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Sulfuric acid, concentrated (> 51% and < 100%) source: ICSC, accessed: 2019-09-04, H2SO4 source: wikidata, accessed: 2019-09-07

Appearance: Clear, colorless liquid source: wikipedia, accessed: 2019-09-27; Odourless colourless oily hygroscopic liquid source: ICSC, accessed: 2019-09-04; Colorless to dark-brown, oily, odorless liquid. [Note: Pure compound is a solid below 51°F. Often used in an aqueous solution.] source: NIOSH NPG, accessed: 2019-09-02

Products:

ZnSO4 – Zinc sulfate source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04

Other names: White vitriol source: wikipedia, accessed: 2019-09-27, Goslarite source: wikipedia, accessed: 2019-09-27, Zinc sulfate (1:1) source: wikidata, accessed: 2019-09-02

Appearance: White powder source: wikipedia, accessed: 2019-09-27; Colourless hygroscopic crystals source: ICSC, accessed: 2019-09-04

Names: Dihydrogen source: wikidata, accessed: 2019-09-07, Hydrogen source: ICSC, accessed: 2019-09-04source: wikidata, accessed: 2019-09-07, H2 source: wikidata, accessed: 2019-09-07

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3 years ago