All categories

3 years ago

Suppose that you want to compare the mass of a block of ice to its mass as liquid

Chemistry

1 answer:

Morgarella [4.7K]3 years ago

5 0

Answer:

You will find the mass of the pan and water but if the water got to its boiling temperature that mass may be a little bit off seeing as some of it may have evaporated

You might be interested in

How much heat is required to raise the temperature of 835g of water from 35.0°C to 65.0°C?

Anna007 [38]

Answer:

25,050 calories.

Explanation:

A calorie is the amount of energy needed to raise the temperature of one gram of water 1 degree centigrade. If we are raising 835 grams of water 30 degrees then we multiply 835*30 to get 25,050 calories.

3 0

3 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago

Convert 0.185mol sucrose (C12H22O11) into grams.

viktelen [127]

1) Find the molar mass of sucrose: 12*12 + 22*1 + 11*16 = 342 g/mol

2) Multiply the molar mass by the number of mols: 0.185 mol * 342 g/mol = 63.27 g

3 0

3 years ago

Identify the number of core and valence electrons for each atom.

AlexFokin [52]

Answer:

1. Core electrons = 1s²; Valence electrons = 2s² and 2p³

2. Core electrons = 1s², 2s²,m2p⁶; Valence electrons = 3s² and 3p⁵

Explanation:

The electrons in an atom of an element are generally divided into two groups: the core electrons and the Valence electrons

Valence electrons are those electrons which are located of or found in the outermost shell or highest energy level of an atom. These valence electrons are the one that determine the chemical reactivity of an atom as they can participate in bond formation between atoms of the same element or with atoms of other elements. The valence electrons also, generally determine the group to which an element will belong to.

Core electrons on the other hand, are those electrons which are found within the innermost shell or lowest energy levels of the atom. They are strongly attracted to the nucleus of atom and do not take part in chemical bonding. However, they play the important role of shieldingnthe valence electrons from the positive charge of the nucleus thereby assisting the valence electrons in bond formation.

3 0

3 years ago

Is first order in bro3⎻ , second order in br⎻, and zero order in h+. By what factor will the reaction rate change if the concent

nata0808 [166]

For a general reaction,

$A+B\rightarrow C$

General expression for rate law will be:

$r=k[A]^{a}[B]^{b}$

Here, r is rate of the reaction, k is rate constant, a is order with respect to reactant A and b is order with respect to reactant B.

The reaction is first order with respect to $BrO_{3}^{-}$ , second order with respect to $Br^{-}$ and zero order with respect to $H^{+}$ .

According to above information, expression for rate law will be:

$r=k[BrO_{3}^{-}]^{1}[Br^{-}]^{2}[H^{+}]^{0}$

Or,

$r=k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (1)

When concentration of $BrO_{3}^{-}$ get doubled, rate of the reaction becomes,

$r^{'}=2k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (2)

Dividing (2) by (1)

$\frac{r^{'}}{r}=\frac{2k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=2$

Or,

$r^{'}=2r$

Thus, rate of the reaction also get doubled.

When the concentration of $Br^{-}$ is halved, the rate of reaction becomes

$r^{"}=k[BrO_{3}^{-}]([Br^{-}]/2)^{2}$

Or,

$r^{"}=1/4k[BrO_{3}^{-}][Br^{-}]^{2}$ ...... (3)

Dividing (3) by (1)

$\frac{r^{"}}{r}=\frac{1/4k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=\frac{1}{4}$

Or,

$r^{"}=\frac{r}{4}$

Thus, rate of reaction becomes 1/4th of the initial rate.

When the concentration of $H^{+}$ is tripled:

Since, the rate expression does not have concentration of $H^{+}$ , it is independent of it. Thus, any change in the concentration will not affect the rate of reaction and rate of reaction remains the same as in equation (1).

7 0

4 years ago