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3 years ago

Mercury(II) oxide (HgO) decomposes to form mercury (Hg) and oxygen (O2). The balanced chemical equation is shown below.

Chemistry

2 answers:

Zina [86]3 years ago

8 0

Answer:

15.63

Explanation:

i just took it on edgeunity

zavuch27 [327]3 years ago

7 0

15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.

Step 1. Convert grams of O_2 to moles of O_2

Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2

Step 2. Use the molar ratio of HgO:O_2 to convert moles of O_2 to moles of HgO

Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = 15.63 mol HgO

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zzz [600]

Answer:

10/24

Simplified: 5/12

Explanation:

Well first, let's convert years into months

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In fraction form we would write this as 10/24

If the question asks for it to be simplified, then simply

Both 10 and 24 evenly fit 2

Therefore, divide both the numerator and the denominator by 2

10 ÷ 2 / 24 ÷ 2

5/12

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2 years ago

A buffer is prepared by adding 300. 0 ml of 2. 0 mnaoh to 500. 0 ml of 2. 0 mch3cooh. what is the ph of this buffer? ka= 1. 8 10

Anton [14]

The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.

<h3>What is the Henderson-Hasselbalch equation?</h3>

Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.

Given,

Acid disassociation constant (ka) = 1. 8 10⁻⁵

Concentration of NaOH = 2.0 M

Concentration of CH₃COOH = 2.0 M

pKa value is calculated as,

pKa = -log Ka

pKa = - log (1. 8 x 10⁻⁵)

Substituting the value of pKa in the Henderson-Hasselbalch equation as

pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]

pH = - log (1. 8 x 10⁻⁵) + log [1]

= 4.745 + 0

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Learn more about the Henderson-Hasselbalch equation here:

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1 year ago

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dolphi86 [110]

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2 years ago

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

Dmitry [639]

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

$4.9 g \times \frac{1mol}{98.00g} = 0.050mol$

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

$0.050mol \times \frac{149.09 g}{mol} = 7.5 g$

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2 years ago