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3 years ago

Aqueous solutions of copper (II) bromide and silver (1) acetate react to form solid

Chemistry

1 answer:

Sladkaya [172]3 years ago

4 0

Answer:

5.63 g

Explanation:

Step 1: Write the balanced equation

CuBr₂(aq) + 2 AgCH₃CO₂(aq) ⇒ 2 AgBr(s) + Cu(CH₃CO₂)₂(aq)

Step 2: Calculate the reacting moles of copper (II) bromide

30.0 mL of 0.499 M CuBr₂ react. The reacting moles of CuBr₂ are:

$0.0300L \times \frac{0.499mol}{L} = 0.0150mol$

Step 3: Calculate the moles formed of silver (I) bromide

The molar ratio of CuBr₂ to AgBr is 1:2. The moles formed of AgBr are 2/1 × 0.0150 mol = 0.0300 mol.

Step 4: Calculate the mass corresponding to 0.0300 mol of AgBr

The molar mass of AgBr is 187.77 g/mol.

$0.0300 mol \times \frac{187.77g}{mol} = 5.63 g$

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Which of the following is not true of acids?

NeTakaya

Answer is: 2. dillute acids feel slipper.

1) Acids are corrosive is correct. For example hydrochloric acid (HCl) will react with most metals.

2) Dillute acids feel slippery is not correct. Bases, for example solution of sodium hydroxide feels slipery.

3) Acids have a distinctly sour taste is correct. For example, vinegar is mixture of acetic acid (CH₃COOH) and water (H₂O). Vinegar is colourless liquid with sour taste and pungent smell, freezing point of the vinegar is lower than glacial acetic acid.

4) Acids have more hydronium ions than hydroxide ions is correct. Because acid gives a lot of hydrogen cations (H⁺), pH (pH = -log[H⁺]) is lower than seven (acidic solution).

5 0

3 years ago

What mass of ammonia can be produced if 13.4 grams of nitrogen gas reacted ?

aivan3 [116]

Answer:

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 13.4 grams

Molar mass of N2 = 28 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles of N2

Moles N2 = Mass N2 / molar mass N2

Moles N2 = 13.4 grams / 28.00 g/mol

Moles N2 = 0.479 moles

Step 4: Calculate moles of NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles

Step 5: Calculate mass of NH3

Mass of NH3 = moles NH3 * molar mass NH3

Mass NH3 = 0.958 moles * 17.03 g/mol

Mass NH3 = 16.3 grams

If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia

3 0

3 years ago

1. A 225-L barrel of white wine has an initial free SO2 concentration of 22 ppm and a pH of 3.70. How much SO2 (in grams) should

Alexandra [31]

Answer:

The appropriate answer is "9.225 g".

Explanation:

Given:

Required level,

= 63 ppm

Initial concentration,

= 22 ppm

Now,

The amount of free SO₂ will be:

= $Required \ level -Initial \ concentration$

= $63-22$

= $41 \ ppm$

The amount of free SO₂ to be added will be:

= $41\times 225$

= $9225 \ mg$

∵ 1000 mg = 1 g

So,

= $9225\times \frac{1}{1000}$

= $9.225$

Thus,

"9.225 g" should be added.

3 0

3 years ago

2CO(g)+O2 (g) =2CO2(g)

Ronch [10]

Answer:

6.76 moles.

Explanation:

2CO(g)+O2 (g) =2CO2(g)

When 2 CO mols were reacted with excess O2 then 2 mols of CO2 is created.

Therefore if 6.76 moles reacted, same number of CO2 will be created.

4 0

3 years ago

Balance the equation<br> N2O5 + H2 -> NH3 + H2O

Lubov Fominskaja [6]

Answer:

I hope this is it. I'm not really sure.

4 0

3 years ago