Answer:
53.7 grams of HNO3 will be produced
Explanation:
Step 1: Data given
Mass of NO2 = 59.0 grams
Molar mass NO2 = 46.0 g/mol
Step 2: The balanced equation
3NO2 + H2O→ 2HNO3 + NO
Step 3: Calculate moles NO2
Moles NO2 = 59.0 grams / 46.0 g/mol
Moles NO2 = 1.28 moles
Step 4: Calculate moles HNO3
For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO
For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3
Step 7: Calculate mass HNO3
Mass HNO3 = 0.853 moles * 63.01 g/mol
Mass HNO3 = 53.7 grams
53.7 grams of HNO3 will be produced
Answer: 116g/mole
Explanation:
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You need evidence to support a conclustion
Answer:
A PROFESSOR WARNED HIM NOT TO GO INTO PHYSICS. Not long after the 16-year-old Planck got to the University of Munich in 1874, physics professor Philipp von Jolly tried to dissuade the young student from going into theoretical physics. Jolly argued that other scientists had basically figured out all there was to know.
Explanation:
<u>Answer:</u>
<u>For a:</u> The empirical formula of the compound is 
<u>For b:</u> The empirical formula of the compound is 
<u>Explanation:</u>
We are given:
Percentage of P = 43.6 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.406 moles.
For Phosphorus = 
For Oxygen = 
Converting the moles in whole number ratio by multiplying it by '2', we get:
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of P : O = 2 : 5
Hence, the empirical formula for the given compound is 
We are given:
Percentage of K = 28.7 %
Percentage of H = 1.5 %
Percentage of P = 22.8 %
Percentage of O = 56.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of K = 28.7 g
Mass of H = 1.5 g
Mass of P = 43.6 g
Mass of O = 56.4 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Potassium =
Moles of Hydrogen =
Moles of Phosphorus =
Moles of Oxygen = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.735 moles.
For Potassium = 
For Hydrogen = 
For Phosphorus = 
For Oxygen = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of K : H : P : O = 1 : 2 : 1 : 4
Hence, the empirical formula for the given compound is 