What inference can be drawn from the graph?

Lead metal is produced by heating solid lead (II) sulfide with solid lead (II) sulfate, resulting in liquid lead and sulfur diox

olya-2409 [2.1K]

Answer:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

Explanation:

The balanced reaction of heating solid lead (II) sulfide with the solid lead (II) sulfate to produce liquid lead and sulfur dioxide gas is shown below:

$PbS_{(s)}+PbSO_4_{(s)}\overset{\Delta}{\rightarrow} 2Pb_{(l)}+2SO_2_{(g)}$

In the balance reaction above, all the phases are indicated. This reaction is used for the production of lead metal.

4 0

3 years ago

What is the product(s) of the reaction below?

sladkih [1.3K]

Answer:

A. Solid aluminum oxide and solid iron

Explanation:

The reaction equation is given as:

2Al + Fe₂O₃ → Al₂O₃ + 2Fe

The species on the left hand side are the reactants

Those on the right hand side of the expression are the products.

The products are:

Al₂O₃ and Fe

These are solid aluminum oxide and solid iron

3 0

3 years ago

Ammonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with

ankoles [38]

Answer:

$n_{NH_3}=0.18molNH_3$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$H_3PO_4(aq)+3NH_3(l)\rightarrow (NH_4)_3PO_4$

In such a way, considering the 3 to 1 molar relationship between ammonia and ammonium phosphate, the moles of ammonia result:

$n_{NH_3}=0.060mol(NH_4)_3PO_4*\frac{3molNH_3}{1mol(NH_4)_3PO_4}\\n_{NH_3}=0.18molNH_3$

Best regards.

8 0

3 years ago

You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30

dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula: $\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n$

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = (\dfrac{1}{3})^4$

$\dfrac{(final \ mass \ of \ solute)_{water}}{0.30 \ g} = 0.012345$

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0

3 years ago

1. The molarity of a solution of 6.0 g of KCl in 80. mL of solution is ________

Ahat [919]

Answer:

1.0 M of KCl

Explanation:

First off, we're given 6.0 g of KCl. From here, we just find the molar mass of KCl by adding the molar masses of each element in the compound. The molar mass for the potassium is 39 g/mol, and the chlorine is 34 g/mol (all found on periodic table). We add those to find the compounds molar mass, which turns out to be 73 grams/ mole.

From here, we need to convert grams to moles so we multiply the grams by the molar mass so that we cancel out grams and only leave behind moles ( 6.0 grams x 1 mole/73 grams ≈ 0.08 moles).

From here, we need to find the amount of liters in the solution, considering that the Molarity (M) is the same equation as mol/ Liter. To find this, we must convert similar units so that they work, in this case, we must convert milliliters to liters. We know that 1000 milliliters are equal to 1 liter, therefore like what we did with the molar mass and grams, we must cancel out milliliter so we only have liters ( 80 mL x 1 L / 1000 mL= 0.08 L)

From here, we divide the moles by the liters in order to make the molarity. The reason why is because as said before, M is equal to mol/L

(0.08 moles / 0.08 L = 1.0 M)

I hope this helped you and good luck with your semester in Chemistry!

7 0

3 years ago