How many moles of sulfur must be burned to give 0.567 moles of so2?

How many moles of sulfur must be burned to give 0.567 moles of so2?
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Description : (3) Oxides of nitrogen and sulphur Explanation: Acid rain is caused by emissions of oxides of Sulfur and Nitrogen (Sulfur Dioxide and Nitrogen Oxide), which react with the water molecules in the ... The emissions of sulfur dioxide (SO2) and nitrogen oxides (NOx) result from fossil fuel combustion.

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Last Answer : 1,8 of moles of O2 are required.

How many moles of sulfur are in 53.7 grams of sulfur?

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Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \frac{3}{4} \)b) \( \frac{5}{8} \)c) \( \frac{9}{8} \)d) \( \frac{7}{8} \)47. The sum of perimeter of all triangle is (in \( cm \) )a) 144b) 625c) 400d) 16948. The area of all the triangle is (in sq cm)a) 576b) \( 144 \sqrt{3} \)c) \( 169 \sqrt{3} \)d) \( 192 \sqrt{3} \)49. The sum of perimeter of first 6 triangle is (in \( cm \) )a) 120b) \( \frac{567}{4} \)c) \( \frac{569}{4} \)d) 14450. The side of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

Description : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \ ... of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3

Last Answer : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla ... ( 1.5 \) c) \( 0.75 \) d) 3

A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

Description : A fluid flows in a steady manner between two section in a flow line at section 1: A 1 = 1ft², V1 = 100fpm, volume1 of 4ft³/lb. at sec2: A2 = 2 ft², p= 0.20 lb/ft³ calculate the velocity at section 2.  a. 625 fpm  b. 567 fpm  c. 356 fpm  d. None of the above

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(37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? (Take 3 digits after the decimal point) a) 95.357 b) 69.567 c) 96.257 d) 78.584 e) 88.598

Description : (37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? (Take 3 digits after the decimal point) a) 95.357 b) 69.567 c) 96.257 d) 78.584 e) 88.598

Last Answer : (37) 0.75 x [(7056) 0.25] ÷ (64) – 2 = (8) ? Sol: 27.75 x 1764 ÷ 64 – 2 = 8 x ? 27.75 x 27.5625 – 2 = 8 x ? 762.859/8=? ? = 95.357 Answer: a)

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Last Answer : Let the Deepa Monthly Salary = x Deepa decided to donate 16% of her monthly salary. On the day of donation she changed. So, 75%of she had decided earlier of 16% 16 × 75/100 = 12% Therefore = 12%= Rs. 6567 100% = Rs .6567 × 100 / 12 x = Rs. 54725 Answer: a)

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