Ask question

Ask question

All categories

2 years ago

6

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!

You might be interested in

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

vovikov84 [41]

Answer:

$\rm C_2H_8N_2O_3$ .

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

$\rm C$ : approximately $12$ .
$\rm H$ : approximately $1$ .
$\rm N$ : approximately $14$ .
$\rm O$ : approximately $16$ .

The relative atomic mass of an element is numerically equal to the mass (in grams, $\rm g$ ,) of one mole of atoms of this element.

For example, the relative atomic mass of $\rm C$ is approximately $12$ . Therefore, each mole of $\rm C\!$ atoms would have a mass of $12\; \rm g$ .

This sample contains $24\; \rm g$ of carbon. That would correspond to approximately $\displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol$ of $\rm C$ atoms.

Similarly, for the other three elements:

$\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol$ .

$\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol$ .

$\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol$ .

Hence, the ratio between these elements in this compound would be:

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ .

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

$n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3$ is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, $\rm C$ , is present in this compound, then:

$\rm C$ (carbon) and then $\rm H$ (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
The other elements in this compound will be listed in alphabetical order.

If there is no carbon $\rm C$ in this compound, then list all the elements in this compound in alphabetical order.

Both $\rm C$ (carbon) and $\rm H$ (hydrogen) are found in this compound. Therefore, the first element in the list would be $\rm C\!$ . The second would be $\rm H\!$ , followed by $\rm N\!$ and then $\rm O\!$ .

Hence, the empirical formula of this compound would be $\rm C_2H_8N_2O_3$ .

6 0

2 years ago

Which of these is a difference between a single skin cell and the skin tissue that covers an entire hand?

fgiga [73]

Answer:

c

Explanation:

the cells grow more than the tissue

7 0

2 years ago

Read 2 more answers

The combustion of propane gas is used to fuel barbeque grills. If 4.65 moles of propane, C3H8, are burned in a grilling session,

Varvara68 [4.7K]

Answer:

<h3>moles of carbon dioxide=13.95mol</h3>

Explanation:

First wrie down the balance chemical reaction:

$C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l)$

Combustion reaction: The reacion in which hydrocarbon is burnt in the presence of oxygen gas and it releases heat and this reaction exothermic because heat of cumbustion is negative.

eg. burning of methane

By using unitry method,

From the above balanced reaction it is clearly that,

from 1 mole of propane 3 moles of carbon dioxide is formed

there fore,

from 4.65 mole of propane $3 \times 4.65$ moles of carbon dioxide will form

moles of carbon dioxide=13.95mol

8 0

2 years ago

A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl

Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

$Kr+2Cl_2\rightarrow KrCl_4$

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, $n=\frac{0.500*15.0}{0.0821*623.8}$

n = 0.146 moles

for chlorine, $n=\frac{1.50*15.0}{0.0821*623.8}$

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

$0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})$

= 32.94 g of $KrCl_4$

So, 32.94 g of the product will form.

5 0

2 years ago

In an ionic bond:

inna [77]

Correct answer is
<span>D. One atom accepts electrons from another.</span>

8 0

2 years ago

Read 2 more answers