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Gekata [30.6K]
3 years ago
6

Phosphoric acid, which is commonly used as rust inhibitor, food additive and etching agent for dental and orthopedic use, can be

synthesized using a two-step thermal process. in the first step, phosphorus and oxygen react to form diphosphorus pentoxide: (l) (g) (g)
Chemistry
1 answer:
igor_vitrenko [27]3 years ago
6 0

The first step in the synthesis of phosphoric acid is the reaction of phosphorus and oxygen to form diphosphorus pentoxide.

P_{4}(l) + 5 O_{2}(g)--->2 P_{2}O_{5} (g)

The second step in the synthesis of phosphoric acid is the reaction of diphosphorus with water to form phosphoric acid.

P_{2}O_{5}(g) + 3H_{2}O (l) --> 2H_{3}PO_{4}(aq)

The net chemical equation representing the formation of phosphoric acid from phosphorus, water and oxygen will be,

Step -1 : P_{4}(l) + 5O_{2}(g) ---> 2P_{2}O_{5}(g)

Step 2: P_{2}O_{5}(g) + 3 H_{2}O(l)-->2 H_{3}PO_{4}(aq)

Equation in step 2 must be multiplied by 2 in order to cancel diphosphorus pentoxide.

P_{4}(l) + 5O_{2}(g)---> 2P_{2}O_{5}

2 [P_{2}O_{5} (g) + 3 H_{2}O(l) --> 2 H_{3}PO_{4} (aq) ]

-----------------------------------------------------------------------------------------------------

Net: P_{4}(l) + 5 O_{2}(g) + 6 H_{2}O(l) --->   4H_{3}PO_{4}(aq)


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At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
Why do you think most lampshades are made of translucent materials instead of transparent materials?
Gala2k [10]

Explanation:

because translucent shades lets the light through easily (gentle diffusion)

6 0
2 years ago
What is the molarity of 50.84 g of Na2CO3 dissolved in 0.400 L solution?
borishaifa [10]

Answer:

1.20 M

Explanation:

Convert grams of Na₂CO₃ to moles.  (50.84 g)/(105.99 g/mol) = 0.4797 mol

Molarity is (moles of solute)/(liters of solvent) = (0.4797 mol)/(0.400 L) = 1.20 M

4 0
3 years ago
Can anyone check my work and see if it is correct? If not, may someone help me?
shusha [124]
It looks all correct to me, great job!
6 0
3 years ago
Some people must eat a low-sodium diet with no more than 2,000 mg of sodium per day. By eating 1 cracker, 1 pretzel, and 1 cooki
Leni [432]

Answer:

The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

Explanation:

Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.

So, the following three equations can be written as per given information:

x+y+z = 149 ........(1)

8y+8z = 936 ........(2)

6x+7y = 535 .........(3)

From equation- (2), y+z = \frac{936}{8} = 117

By substituting the value of (y+z) in equation- (1) we get,

                          x = 149-(y+z) = 149-117 = 32

By substituting the value of x into equation- (3) we get,

                           y = \frac{535-(6\times 32)}{7} = 49

By substituting the value of y  into equation- (2) we get,

                           z = (117-49) = 68

So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.

6 0
3 years ago
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