2 points

At 1023 K and 1 atm, a 3.00 gram sample of Snoz(s) (gram-formula mass = 151 g/mol) reacts with hydrogen gas to produce tin and w

Blizzard [7]

Answer:

2 moles of Sn are produced when 4 moles of H2(g) are consumed completely

Explanation:

to determine the number of moles of sn (l) produced when 4.0 moles of H2 (g) is consumed completely.

First, find the number of moles of H2 consumed by taking this as limiting reagent.

$n = \frac{g}{M.W (g/mol)}$

Then find the moles of Sn (l) taking into account the stoichiometric relationship between H2(g) and Sn(l). 2:1

$SnO_{2}$ (s) + 2 $H_{2}$ (g) ⇒ Sn(l) + 2 $H_{2}O$ (g)

$mol Sn(l) = \frac{1mol Sn}{2mol H_{2} } . 4 mol H_{2} = 2 mol$

∴2 moles of Sn are produced when 4 moles of H2(g) are consumed completely.

4 0

2 years ago

Calculate the magnitude and direction (i.e., the angle with respect to the positive H-axis, measured positive as counter-clockwi

r-ruslan [8.4K]

The graphics in the attachment is part of the question, which was incomplete.

Answer: Fr = 102N and angle of approximately 11°.

Explanation: From the attachment, it is observed that from the three forces acting on M, two are perpendicular. So to find them, we have to show their x- and y- axis components. From the graph:

Fx = 70+40-10 = 100

Fy = 40-20 = 20

Now, as the forces form a triangle, the totalforce is:

Fr = $\sqrt{Fx^{2} +Fy^{2} }$

Fr = $\sqrt{10400}$

Fr = ≈ 102N

To determine the angle requested, we use:

arctg H = $\frac{Fy}{Fx}$

arctg H = $\frac{20}{100}$

H = tg 0.2 ≈ 11°.

4 0

3 years ago

The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and

Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

Moles KSCN = Moles Ag⁺ in the filtrate:

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

Total moles Ag⁺:

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

Moles of Ag⁺ in the precipitate:

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

Moles AsO₄ = Moles As:

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

Moles As₂O₃:

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

Mass As₂O₃:

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

7 0

3 years ago

Which term best describes what Donna observed? A:dew B:fog C:mist D:rain

MariettaO [177]

I also think it’s B but not quite sure

6 0

3 years ago

If 10.00 g of iron metal is burned in the presence of excess of O2 how many grams of Fe2O3 will form

sergiy2304 [10]

14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.

Explanation:

The balanced equation for the reaction is to be known so that number of moles taking part can be known.

The balanced chemical equation is

4Fe + 3 $O_{2}$ ⇒ 2 $Fe{2}$ $O{3}$

From the given weight of iron to be used for the production of $Fe{2}$ $O{3}$ , number of moles of Fe taking part in the reaction can be known by the formula:

Number of moles= mass ÷ Atomic mass of one mole of the element.

(Atomic weight of Fe is 55.845 gm/mole)

Putting the values in equation

Number of moles = 10 gm ÷ 55.845 gm/mole

= 0.179 moles

Applying the stoichiometry concept

4 moles of Fe gives 2 Moles of Fe2O3

0.179 moles will produce x moles of Fe2O3

So, 2÷ 4 = x ÷ 0.179

2/4 = x/ 0.179

2 × 0.179 = 4x

2 × 0.179 / 4 = x

x = 0.0895 moles

So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.

Now the formula used above will give the weight of Fe2O3

weight = atomic weight × number of moles

= 159.69 grams × 0.0895

= 14.292 grams of Fe2O3 formed.

4 0

3 years ago