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2 years ago

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Insoluble sulfide compounds are generally black in color. Which of the following combinations could yield a black precipitate? C

heck all that apply. Na2S(aq) + KCl(aq) Li2S(aq) + Pb(N03)2(aq) Pb(C103)2(aq) + NaNO3(aq) Try Again; 3attempts remaining AgNo3(aq) + KCl(aq) K2S(aq) + Sn(N03)4(aq)

1 answer:

Lubov Fominskaja [6]2 years ago

6 0

Answer:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

Explanation:

For these reactions, the products are:

1. Na₂S(aq) + 2 KCl(aq) → K₂S + 2 NaCl

2. Li₂S(aq) + Pb(NO₃)₂(aq) → PbS + 2 Li(NO₃)

3. Pb(ClO₃)₂(aq) + 2 NaNO₃(aq) → Pb(NO₃)₂ + 2 Na(ClO₃)

4. AgNO₃(aq) + KCl(aq) → AgCl + KNO₃

5. 2 K₂S(aq) + Sn(NO₃)₄(aq) → SnS₂ + 4 KNO₃

K₂S is a soluble sulfide that, in solid state, is white. As K has a low electronegativity, the relative polar bond K-S allows its dissolution in water.

PbS is a black and insoluble compound. As Pb electronegativity is higher than K electronegativity, The bond is less-polar and its dissolution will not be allowed.

For the third reaction, all nitrates are soluble and Na(ClO₃) is also a very soluble compound.

AgCl is an insoluble white compound.

As Sn electronegativity is high, the solubility of this compound is very low. Also, this compound is black!

Thus, combinations could yield a black precipitate are:

Li₂S(aq) + Pb(NO₃)₂(aq)

2 K₂S(aq) + Sn(NO₃)₄(aq)

I hope it helps!

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The Safe Drinking Water Act (SDWA) sets a limit for mercury-a toxin to the central nervous system-at 0.002 mg/L. Water suppliers

posledela

Answer:

The volume of mercury-contaminated water that has to be consumed to ingest 0.100 g mercury is 2.50 × 10⁴ l

Explanation:

Hi there!

First, let´s convert 0.100 g to mg:

0.100 g · (1000 mg/1 g) = 100 mg

The contaminated water has 0.004 mg per liter, then, we have to find the volume of water that contains 100 mg of mercury:

100 mg · (1 l / 0.004 mg) = 2.50 × 10⁴ l

Then, the volume of mercury-contaminated water ( at a concentration of 0.004 mg/l) that has to be consumed to ingest 0.100 g mercury is 2.50 × 10⁴ l

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2 years ago

A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)

allochka39001 [22]

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

Atomic mass = 197 g

then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

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h = 7159.2 cm ≈ 7160 cm

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2 years ago

Read 2 more answers

An amount of solid barium chloride, 20.8 g, is dissolved in 100 g water in a coffee-cup calorimeter by the reaction: BaCl2 (s) 

mamaluj [8]

Answer : The enthalpy change during the reaction is -6.48 kJ/mole

Explanation :

First we have to calculate the heat gained by the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

m = mass of water = 100 g

c = specific heat = $4.04J/g^oC$

$T_{final}$ = final temperature = $26.6^oC$

$T_{initial}$ = initial temperature = $25.0^oC$

Now put all the given values in the above formula, we get:

$q=100g\times 4.04J/g^oC\times (26.6-25.0)^oC$

$q=646.4J$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles barium chloride = $\frac{\text{Mass of barium chloride}}{\text{Molar mass of barium chloride}}=\frac{20.8g}{208.23g/mol}=0.0998mole$

$\Delta H=-\frac{646.4J}{0.0998mole}=-6476.95J/mole=-6.48kJ/mole$

Therefore, the enthalpy change during the reaction is -6.48 kJ/mole

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2 years ago

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$

$\\ 1.22 * 10^{5} = -19953.6 * ln K$

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So, the equilibrium constant is 0.0022.

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2 years ago

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction: C(s)+2H2O(g)→2H2

madam [21]

The question is incomplete , complete question is:

Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.

Answer:

The ΔH of the reaction is -626 kJ/mol.

Explanation:

$C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?$

We are given with:

$\Delta H_{H-O}=459 kJ/mol$

$\Delta H_{H-H}=432 kJ/mol$

$\Delta H_{C=O}=799 kJ/mol$

ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)

$\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})$

$=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol$

$\Delta H=-626 kJ/mol$

The ΔH of the reaction is -626 kJ/mol.

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2 years ago