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tresset_1 [31]
4 years ago
5

In the reaction below, if 5.71 g of sulfur is reacted with 10.0 g of oxygen, how many grams of sulfur trioxide will be produced?

Chemistry
1 answer:
yawa3891 [41]4 years ago
3 0

Answer:

14.3 g SO₃

Explanation:

2S + 3O₂ → 2SO₃

First, find the limiting reactant.  To do that, calculate the mass of oxygen needed to react with all the sulfur.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂

0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂

There are 10.0 g of O₂, so there's enough oxygen.  The limiting reactant is therefore sulfur.

Use the mass of sulfur to calculate the mass of sulfur trioxide.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃

0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃

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DA = 13, BW = 5, WC = 12, ∠ACD = 25°, ∠DAB = 50°, ∠ADC = 130°, ∠DBC = 65° and ∠BWC = 90° given that ABCD is a rhombus and DB = 10, BC = 13, ∠WAD = 25°. This can be obtained by understanding the properties of a rhombus.

<h3>What are the required values of sides and angles?</h3>

Here in the question it is given that,

  • ABCD is a rhombus
  • DB = 10, BC = 13, ∠WAD = 25°

We have to find the values of DA, BW, WC, ∠BAC, ∠ACD, ∠DAB, ∠ADC, ∠DBC, ∠BWC.

1) Since all the sides of the rhombus are equal, DA = 13

2) Since diagonals are perpendicular bisectors of each other, BW = 5

3) Since formula of diagonal is,

d₁ = √4a² - d₂,

here a = 13, d₂ = 10

d₁ = √4×13² - 10

d₁ = √676 - 10 = √576 = 24

Since diagonals are perpendicular bisectors of each other

WC = 24/2  ⇒ WC = 12

4) Since diagonals are angle bisectors at the corners, ∠BAC = 25°

5) Since diagonals are angle bisectors at the corners,

∠BAW = ∠DAW = 25° ⇒ ∠BAD =∠BAW + ∠DAW = 25° + 25° ⇒∠BAD = 50°

Since opposite angles of the rhombus are equal,

∠BAD = ∠BCD = 50° ⇒ ∠BCW = ∠DCW = 25°

⇒ ∠ACD = 25°

6) Since diagonals are angle bisectors at the corners,∠BAW = ∠DAW = 25° ⇒ ∠BAD =∠BAW + ∠DAW = 25° + 25° ⇒∠BAD = 50°

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7) Since adjacent angles are supplementary angles,

∠BAD + ∠ADC = 180° ⇒ 50° + ∠ADC = 180° ⇒ ∠ADC = 180° - 50° = 130°

⇒ ∠ADC = 130°

8) Since diagonals are angle bisectors at the corners,

∠DBC = 130°/2

⇒ ∠DBC = 65°

9) Since all the sides make an angle of 90° at the center,

⇒ ∠BWC = 90°

Hence DA = 13, BW = 5, WC = 12, ∠ACD = 25°, ∠DAB = 50°, ∠ADC = 130°, ∠DBC = 65° and ∠BWC = 90° given that ABCD is a rhombus and DB = 10, BC = 13, ∠WAD = 25°.

Learn more about rhombus here:

brainly.com/question/27870968

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For a straight line graph, the normal equation could be written as

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Writing the equation is the form of y = mx + c gives the following;

lnN = -kt + No

As obtained from the question, t is plotted in the x-axis, while lnN is the plot on the y-axis.

Now, we are asked to obtain the slope and the y-intercept. It can be seen that the slope is the coefficient of the term x in this case t. Hence, our slope is -k.

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